剑指&leetcode刷题记录

class Solution {
public:
    int jumpFloorII(int number) 
    {
        return 1<<(number-1);
    }
};

题目：变态跳台阶

题目讲解：https://www.nowcoder.com/questionTerminal/22243d016f6b47f2a6928b4313c85387

 

 

题目：最小的K个数问题

class Solution {
public:
    vector GetLeastNumbers_Solution(vector input, int k) 
    {
        vector result;
        priority_queue, greater >  q;//greater<>是仿函数，用于申明小根堆
        int size = input.size();
        if(size < k) result;
        for(int i=0;i

本题还有个最优解：利用bfprt算法

 

题目 4：替换空格

解法1：用了额外空间的O（N）算法

class Solution {
public:
	void replaceSpace(char *str,int length) 
    {
        string s;
        char *p = str;
        while(*p != '\0')
        {
            if(*p != ' ')
                s += *p;
            else s += "%20";
            p++;
        }
        int i = 0;
        for(; i

解法2：不用额外空间的O(N) 的算法

class Solution {
public:
	void replaceSpace(char *str,int length) 
    {
        int size = 0;
        if(str == NULL || length <=0)
            return ;
        int count = 0;
        while(str[size] != '\0')
        {
            if(str[size] == ' ') count ++;
            size++;
        }
        --size;
        int newLen = size + 2*count;
        if(newLen > length) return ;
        
        int indexOne = size;
        int indexTwo = newLen;
        
        while(indexOne>=0 && indexOne

题目5：从尾到头打印链表

class Solution {
public:
    vector printListFromTailToHead(ListNode* head) 
    {
        stack myStack;
        vector result;
        ListNode* n = head ;
        if(n == NULL)
            return result;
        while(n)
        {
            myStack.push(n->val);
            n = n->next;
        }
        while(!myStack.empty())
        {
            result.push_back(myStack.top());
            myStack.pop();
        }
        return result;
    }
};

经典的栈（递归）结构

 

面试题7：用两个栈实现队列

class Solution
{
public:
    void push(int node) 
    {
        stack1.push(node);
    }

    int pop() 
    {
        if(stack2.empty())
        {
            while(!stack1.empty())
            {
                stack2.push(stack1.top());
                stack1.pop();
            }
        }
        if(stack2.empty()) return -1;
        else
        {
            int result = stack2.top();
            stack2.pop();
            return result;
        }
    }

private:
    stack stack1;
    stack stack2;
};

leetcode:33. 搜索旋转排序数组

class Solution {
public:
    int fun(vector& nums,int target,int begin,int end)
    {
        if(begin < 0 || end >=nums.size())
            return -1;
        int midan;
        if(nums[begin] == target) return begin;
        else if(nums[end] == target) return end;
        
        while(begin < end)
        {
            midan = (end+begin)/2;
            if(nums[midan] == target) return midan;
            else if(nums[midan] < target)
                begin = midan+1;
            else if(nums[midan] > target)
                end = midan-1;
        }
        if(begin == end && nums[begin] == target) return begin;
        else return -1;
    }
    int search(vector& nums, int target) 
    {
        int size = nums.size();
        if(size == 0) return -1;
        int indexLeft = 0;
        int indexRight = size-1;
        int indexMidan = indexLeft;
        int indexMin;
        while(nums[indexLeft] > nums[indexRight])
        {
            indexMidan = (indexRight + indexLeft)/2;
            if(indexLeft+1 == indexRight)
                break;
            if(nums[indexLeft] < nums[indexMidan])//中点在左边数组,最小值在右边,不会是自己
                indexLeft = indexMidan;
            else if(nums[indexLeft] > nums[indexMidan])//中点在右边数组，最小值在左边，也可能是自己
                indexRight = indexMidan;
        }
        indexMin = indexRight;
        if(indexMin == 0)
            return fun(nums,target,0,size-1);
        else if(target <= nums[size-1] && target >= nums[indexMin])
            return fun(nums,target,indexMin,size-1);
        else if(target >=nums[0] && target <=nums[indexMin-1])
            return fun(nums,target,0,indexMin-1);
        return -1;
    }
};

leetcode:74. 搜索二维矩阵

log n + log m

class Solution {
public:
    bool searchMatrix(vector>& matrix, int target) 
    {
        int hang = matrix.size();
        if(hang == 0) return false;
        int lie = matrix[0].size();
        if(lie == 0) return false;
        int left = 0;
        int right = hang-1;
        int k = 0;
        int midan;
        while(left+1 target)
            {
                k=left;
                break;
            }
            else if(matrix[left][lie-1] == target)
                return true;
            else if(matrix[left][lie-1] < target)
            {
                if(matrix[midan][lie-1] < target)
                    left = midan;
                else if(matrix[midan][lie-1] == target)
                    return true;
                else if(matrix[midan][lie-1] > target)
                    right = midan;
            }
            
        }
        if(matrix[left][lie-1] < target)
            k=right;
        
        left = 0;right=lie-1;
        while(left

leetcode:191. 位1的个数

class Solution {
public:
    int hammingWeight(uint32_t n) 
    {
        int result = 0;
        while(n)
        {
            n &= (n-1);
            ++result;
        }
        return result;
    }
};

leetcode:136. 只出现一次的数字

class Solution {
public:
    int singleNumber(vector& nums) 
    {
        int result=nums[0];
        int size = nums.size();
        for(int i=1;i

 

leetcode:62. 不同路径

class Solution {
public:
    int uniquePaths(int m, int n) 
    {
        int** arr = new int*[m];
        for(int i=0;i

leetcode:392. 判断子序列

class Solution {
public:
    bool isSubsequence(string s, string t) 
    {
        int ssize = s.size();
        int tsize = t.size();
        int j=0;
        for(int i=0;i

leetcode:922. 按奇偶排序数组 II

class Solution {
public:
    vector sortArrayByParityII(vector& arr) 
    {
        int size = arr.size();
        if(size == 0 && size == 1) return arr;
        int left = 1;
        int right = size - 2;
        while(left=0)
        {
            while(left=0 && (arr[right]&1) == 0)
                right-=2;
            if(left=0)
                swap(arr[left],arr[right]);
        }
        return arr;
    }
};

剑指offer：反转链表

class Solution {
public:
    ListNode* ReverseList(ListNode* pHead) 
    {
        if(pHead == NULL)
            return NULL;
        ListNode* left = NULL;
        ListNode* right = pHead;
        ListNode* m = NULL;
        
        while(right)
        {
            left = m;
            m = right ;
            right = right->next;
            m->next = left;
        }
        return m;
        
        
        
        
        
       /* if(pHead == NULL)
            return NULL;
        ListNode* p = pHead;
        stack myStack;
        while(p->next)
        {
            myStack.push(p);
        }
        pHead = p;
        while(!myStack.empty())
        {
            p->next = myStack.top();
            p = p->next;
            myStack.pop();
        }
        return pHead;*/
    }
};

剑指offer：倒数第k个链表节点

class Solution {
public:
    ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) 
    {
        if(pListHead == NULL)
            return NULL;
        ListNode* left = pListHead;
        ListNode* right = left;
        for(int i=0;inext == NULL)
                return NULL;
            right = right->next;
        }
        
        while(right->next)
        {
            left = left->next;
            right = right->next;
        }
        return left;
    }
};

剑指offer：合并有序链表

class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        if(pHead1 == NULL)
            return pHead2;
        else if(pHead2 == NULL)
            return pHead1;
        
        if(pHead1->val < pHead2->val)
        {
            pHead1->next = Merge(pHead1->next,pHead2);
            return pHead1;
        }
        else if(pHead1->val >= pHead2->val)
        {
            pHead2->next = Merge(pHead1,pHead2->next);
            return pHead2;
        }
        else return NULL;
    }
};

剑指offer：第一个只出现一次的字符

class Solution {
public:
    int FirstNotRepeatingChar(string str) 
    {
        map m;
        int arr[52];
        int size = str.size();
        if(size == 0) return -1;
        for(int i=0;i::iterator it = m.find(str[i]);
            if(it == m.end())
                m.insert(pair(str[i],1));
            else it->second++;
        }
        for(int i=0;i::iterator it = m.find(str[i]);
            if(it == m.end()) continue;
            if(it->second == 1) return i;
        }
        return -1;
    }
};

剑指offer：包含min函数的栈

class Solution {
public:
    stack myStack, minStack;
    
    void push(int value) 
    {
        myStack.push(value);
        if(minStack.empty()) minStack.push(value);
        else
        {
            int tim = minStack.top();
            if(tim <= value) minStack.push(tim);
            else minStack.push(value);
        }
    }
    void pop() 
    {
        if(myStack.empty())
            return;
        myStack.pop();
        minStack.pop();
    }
    int top() 
    {
        if(myStack.empty()) return -1;
        return myStack.top();
    }
    int min() 
    {
        if(minStack.empty())
            return -1;
        return minStack.top();
    }
};

剑指offer：数组中只出现一次的数字

class Solution {
public:
    void FindNumsAppearOnce(vector data,int* num1,int *num2) 
    {
        vector nums1;
        vector nums2;
        int size = data.size();
        if(size < 2) return ;
        
        int tim = data[0];
        for(int i=1;i

leetcode：3. 无重复字符的最长子串

class Solution {
public:
    int lengthOfLongestSubstring(string s) 
    {
        map m;
        int result = 0;
        int size = s.size();
        int left = 0;
        map::iterator it;
        for(int i=0;isecondsecond+1);
                it->second = i;
            }
            
            result = result<(i-left+1)?(i-left+1):result;
        }
        return result;
    }
};

剑指offer：树的子结构

class Solution {
public:
    
    bool fun(TreeNode* p1,TreeNode* p2)
    {
        if(p2 == NULL) return true;
        if(p1 == NULL) return false;
        if(p1->val == p2->val)
            return fun(p1->left,p2->left) && fun(p1->right,p2->right);
        return false;
    }
    
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        bool flag = false;
        if(pRoot1 == NULL || pRoot2 == NULL)
            return false;
        
        if(pRoot1->val == pRoot2->val)
        {
            flag =  fun(pRoot1,pRoot2);
        }
        if(flag == false)
        {
            flag = HasSubtree(pRoot1->left,pRoot2);
        }
        if(flag == false)
        {
            flag = HasSubtree(pRoot1->right,pRoot2);
        }
        return flag;
    }
};

剑指offer：二叉树的镜像

class Solution {
public:
    void Mirror(TreeNode *pRoot) 
    {
        if(pRoot == NULL) return;
        
        TreeNode* tim = pRoot->left;
        pRoot->left = pRoot->right;
        pRoot->right = tim;
        Mirror(pRoot->right);
        Mirror(pRoot->left);
    }
};

剑指offer：栈的压入，弹出序列

class Solution {
public:
    bool IsPopOrder(vector pushV,vector popV) 
    {
        int size1 = pushV.size();
        int size2 = popV.size();
        stack myStack;
        int i = 0;
        int j = 0;
        while(i < size2)
        {
            if(myStack.empty()) myStack.push(pushV[j++]);
            if(popV[i] == myStack.top())
            {
                myStack.pop();
                ++i;
            }
            else if(j < size1)
            {
                myStack.push(pushV[j++]);
            }
            else
            {
               return false; 
            }
        }
        if(myStack.empty() && j == size1 && i==size2) return true;
        else return false;
        
    }
};

剑指offer：从上往下打印二叉树（层次遍历）

class Solution {
public:
    vector PrintFromTopToBottom(TreeNode* root) 
    {
        vector result;
        if(root == NULL) return result;
        queue myQueue;
        myQueue.push(root);
        TreeNode *tim;
        while(!myQueue.empty())
        {
            tim = myQueue.front();
            result.push_back(tim->val);
            if(tim->left != NULL) myQueue.push(tim->left);
            if(tim->right != NULL) myQueue.push(tim->right);
            myQueue.pop();
        }
        return result;
    }
};

剑指offer：二叉搜索树的后序遍历序列

class Solution {
public:
    bool fun(vector& v,int begin,int end)
    {
        int size = v.size();
        if(begin < 0 || end >= size)
        {
            return false;
        }
        if(begin == end) return true;
        int root = end;
        int right = begin;
        for(int i = begin;i sequence) 
    {
        return fun(sequence, 0, sequence.size()-1);
    }
};

leetcode：55.跳跃游戏

class Solution {
public:
    bool  fun(vector& nums, int index)
    {
        int size = nums.size();
        if(size == 0) return false;
        if(index == size-1) return true;
        else if(index > size-1) return false;
        bool result = false;
        for(int i=1;i<=nums[index];++i)
        {
            result |= fun(nums,index+i);
        }
        return result;
    }
    bool canJump(vector& nums) 
    {
        return fun(nums,0);
    }
};

递归版本，leetcode上超时。

 

贪心版本

class Solution {
public:
    bool canJump(vector& nums) 
    {
        int size = nums.size();
        int max = 0;
        for(int i=0;i(nums[i]+i)?max:(nums[i]+i);
        }
        return true;
    }
};

leetcode：23. 合并K个排序链表

class Solution {
public:
    
template
struct greater1
{	
	constexpr bool operator()(const _Ty& _Left, const _Ty& _Right) const
	{	
		return (_Left->val > _Right->val);
	}
};
    
    ListNode* mergeKLists(vector& lists) 
    {
        int size = lists.size();
        if(size == 0) return NULL;
        ListNode* result = NULL;
        ListNode* p = NULL;
        priority_queue, greater1> test;
        
        for(int i=0;inext = top;
                p = p->next;
            }
            
            if(top != NULL && top->next != NULL)
            {
                test.push(top->next);
            }
        }
        return result;
        
    }
};

剑指offer：二叉树中和为某一值的路径

class Solution {
public:
    vector> result;
    void Find(TreeNode* root,int expectNumber,int preSum,vector& myStack)
    {
        if(root == NULL) return ;
        
        preSum+=root->val;
        myStack.push_back(root->val);
        
        if(root != NULL && preSum == expectNumber && root->left == NULL && root->right == NULL)
        {
            result.push_back(myStack);
            myStack.pop_back();
            return ;
        }

        Find(root->left,expectNumber,preSum,myStack);
        Find(root->right,expectNumber,preSum,myStack);
        
        myStack.pop_back();
    }
    vector> FindPath(TreeNode* root,int expectNumber) 
    {
        if(root == NULL)
            return result;
        vector myStack;
        Find(root,expectNumber,0,myStack);
        return result;
        
    }
};

剑指offer:数组中的逆序对

class Solution {
public:
    long mergeSort(vector& nums, int begin, int end)
    {
        long result = 0;
        int size = nums.size();
        if(size == 0 || begin < 0 || end >= size)
            return 0;
        if(begin == end) return 0 ;
        int mid = (end + begin) >> 1;
        result += mergeSort(nums, begin, mid);
        result += mergeSort(nums, mid+1, end);
        vector tim;
        int i = begin;
        int j = mid+1;
        while(i <= mid && j<=end)
        {
            if(nums[i] < nums[j])
            {
                tim.push_back(nums[i++]);
            }
            else 
            {
                result += (mid-i+1);
                tim.push_back(nums[j++]);
            }
        }
        while(i<=mid)
        {
            tim.push_back(nums[i++]);
        }
        while(j<=end)
        {
            tim.push_back(nums[j++]);
        }
        for(int k = 0;k data) 
    {
        return mergeSort(data,0,data.size()-1)%1000000007;
    }
};

剑指offer：连续子数组的最大和

class Solution {
public:
    int FindGreatestSumOfSubArray(vector array) 
    {
        int size = array.size();
        if(size == 0) return 0;
        int maxResult = array[0];
        int tim = 0;
        for(int i=0;i

剑指offer：最小的k个数

class Solution {
public:
    int partition(vector& nums,int begin, int end)
    {
        int size = nums.size();
        if(size == 0 || begin < 0 || end >= size || end < begin)
            return -1;
        
        int index = (rand() % (end-begin+1))+ begin;
        int small = begin - 1;
        swap(nums[index],nums[end]);
        for(int i=begin;i GetLeastNumbers_Solution(vector input, int k) 
    {
        int size = input.size();
        vector result;
        if (size == 0) return result;
        if (k == size)
        {
            return input;
        }
        else if (k > size) return result;
        int begin = 0; int end = size - 1;
        int index = partition(input, begin, end);
        while (index != k)
        {
            if (index == -1) return result;
            else if (index < k)
            {
                begin = index + 1;
            }
            else if (index > k)
            {
                end = index - 1;
            }
            index = partition(input, begin, end);
        }
        int K = 0;
        for (int i = 0; i < size; ++i)
        {
            if (input[i] < input[index])
            {
                result.push_back(input[i]);
                ++K;
            }
        }
        for (; K < k; ++K)
        {
            result.push_back(input[index]);
        }
        return result;
    }
};

剑指offer：数组中出现次数超过一半的数

class Solution {
public:
    int partition(vector nums,int begin, int end)
    {
        int size = nums.size();
        if(size == 0 || begin <0 || end >= size)
            return -1;
        
        int index = (rand() % (end-begin+1))+ begin;
        swap(nums[index],nums[end]);
        
        int small = begin - 1;
        for(int i=begin;i numbers) 
    {
        int size = numbers.size();
        if(size == 0) return 0;
        int mid = size >> 1;
        int index = 0;
        int begin = 0, end = size-1;
        while(index != mid)
        {
            if(index < mid)
            {
                begin = index+1;
            }
            else if(index > mid)
            {
                end = index-1;
            }
            index = partition(numbers,begin,end);
        }
        int count = 0;
        for(int i=0;i mid) return numbers[index];
        else return 0;
        
        
    }
};

剑指offer：两个链表的第一个公共节点

class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) 
    {
        ListNode* result = NULL;
        if(pHead1 == NULL || pHead2 == NULL) return result;
        stack s1,s2;
        ListNode* p = pHead1;
        while(p)
        {
            s1.push(p);
            p = p->next;
        }
        p = pHead2;
        while(p)
        {
            s2.push(p);
            p = p->next;
        }
        while(!s1.empty() && !s2.empty())
        {
            if(s1.top() == s2.top())
            {
                result = s1.top();
                s1.pop();
                s2.pop();
            }
            else
            {
                return result;
            }
        }
        return result;
    }
};

剑指offer：和为s的两个数字

class Solution {
public:
    vector FindNumbersWithSum(vector array,int sum) 
    {
        vector result;
        int size = array.size();
        if(size<2) return result;
        map m;
        
        for(int i=0;i::iterator it = m.find(sum - array[i]);
            if(it != m.end() && it->second != i)
            {
                tim = array[i]*(it->first);
                if(tim < min)
                {
                    min = tim;
                    result.clear();
                    result.push_back(array[i]);
                    result.push_back(it->first);
                }
            }
        }
        return result;
    }
};

剑指offer：二叉树的深度

class Solution {
public:
    int treeDepth(TreeNode* root, int preDepth)
    {
        if(root == NULL) return preDepth;
        int left, right;
        left = treeDepth(root->left,preDepth+1);
        right = treeDepth(root->right,preDepth+1);
        return left>right?left:right;
    }
    int TreeDepth(TreeNode* pRoot)
    {
        return treeDepth(pRoot, 0);
    }
};

剑指offer：左旋转字符串

class Solution {
public:
    string LeftRotateString(string str, int n) 
    {
        int size = str.size();
        if(size <= 1) return str;
        n %= size;
        string s1 = "";
        string s2 = "";
        for(int i = 0;i

剑指offer：翻转单词顺序列

class Solution {
public:
    string ReverseSentence(string str) 
    {
        string s;
        map m;
        int size = str.size();
        if(size == 0) return s;
        int count = 0;
        string k = " ";
        for(int i=0;i=0;--i)
        {
            map::iterator it = m.find(i);
            if(it != m.end())
            {
                if(i != 0)
                    result += it->second + " ";
                else
                    result += it->second;
            }
        }
        return result;
    }
};

剑指offer：滑动窗口的最大值

class Solution {
public:
    vector maxInWindows(const vector& num, unsigned int size)
    {
        deque q;
        vector result;
        int nsize = num.size();
        if(nsize == 0 || size == 0) return result;
        
        for(int i=0;i= size-1) result.push_back(num[q.front()]);
        }
        return result;
    }
};

剑指offer：平衡二叉树

class Solution {
public:
    bool fun(TreeNode* root, int& preD)
    {
        if(root == NULL) 
        {
            preD = 0;
            return  true;
        }
        int left,right;
        bool result = true;
        result &= fun(root->left,left);
        result &= fun(root->right,right);
        preD = left>right?left+1:right+1;
        if(abs(left - right) > 1) return false;
        else return result;
    }
    bool IsBalanced_Solution(TreeNode* pRoot) 
    {
        int i = 0;
        return fun(pRoot,i);
    }
};

剑指offer：和为s的连续正数序列

class Solution {
public:
    vector> FindContinuousSequence(int sum) 
    {
        vector> result;
        if(sum < 3) return result;
        
        int left = 1,right = 2;
        int mid = sum/2+1;
        int tmp = 3;
        vector v;
        while(leftsum)
            {
                tmp-=left;
                ++left;
            }
            else if(tmp == sum)
            {
                v.clear();
                for(int i=left;i<=right;++i) v.push_back(i);
                result.push_back(v);
            }
        }
        return result;
    }
};

剑指offer：不用加减乘除做加法

class Solution {
public:
    int Add(int num1, int num2)
    {
        int n1, n2;
        do
        {
           n1 = num1^num2;
           n2 = (num1&num2)<<1;
            num1 = n1;
            num2 = n2;
        }while(num2);
        return num1;
    }
};

剑指offer:圆圈最后剩下的数

class Solution {
public:
    int LastRemaining_Solution(int n, int m)
    {
        if(n<1 || m<1) return -1;
        if(n == 1) return 0;
        return (LastRemaining_Solution(n-1,m)+m)%n;
    }
};

判断一个数组中哪些可能是partition函数调用后枢纽值

https://pintia.cn/problem-sets/994805260223102976/problems/994805278589960192

vector fun(vector nums)
{
	vector result;
	int size = nums.size();
	if (size <= 1) return nums;
	int leftMax = 0x80000000;
	stack s;
	for (int i = size - 1; i > 0; --i)
	{
		if (s.empty()) s.push(nums[i]);
		else if (s.top() > nums[i]) s.push(nums[i]);
		else s.push(s.top());
	}
	for (int i = 0; i < size; ++i)
	{
		if (nums[i] >= leftMax && (s.empty() || nums[i] <= s.top()))
		{
			result.push_back(nums[i]);
			leftMax = nums[i];
		}
		else
		{
			leftMax = leftMax < nums[i] ? nums[i] : leftMax;
		}
		if(!s.empty()) s.pop();
	}
	return result;
}

剑指offer：数据流中的中位数

class Solution {
public:
    priority_queue,greater> min;
    priority_queue max;
    void Insert(int num)
    {
        int tim = num;
		if (((min.size() + max.size()) & 1) == 0)//偶数
		{
			if (!max.empty() && num < max.top())
			{
				max.push(num);
				tim = max.top();
				max.pop();
			}
			min.push(tim);
		}
		else
		{
			if (!min.empty() && num > min.top())
			{
				min.push(num);
				tim = min.top();
				min.pop();
			}
			max.push(tim);
		}
    }

    double GetMedian()
    { 
        int size = min.size() + max.size();
        if((size & 1) == 1) return (double)min.top();
		else return ((double)max.top() + (double)min.top()) / 2;
    }

};

leetcode：1005. K 次取反后最大化的数组和

class Solution {
public:
    int largestSumAfterKNegations(vector& A, int K) 
    {
        priority_queue,greater> q;
        int size = A.size();
        if(size == 0) return 0;
        for(int i=0;i

leetcode：1007. 行相等的最少多米诺旋转

class Solution {
public:
    int find(int* arr,int size,int length)
    {
        int result = 0;
        int max = arr[0];
        for(int i=0;i= max)
            {
                //if(arr[i] == length) return -1;
                result = i;
                max = arr[i];
            }
        }
        return result;
    }
    int minDominoRotations(vector& A, vector& B)
    {
       int leftMin = -1, rightMin = -1;
        int size = A.size();
        int arrA[7] = { 0,0,0,0,0,0,0 }, arrB[7] = { 0,0,0,0,0,0,0 };

        for (int i = 0; i < size; ++i)
        {
            arrA[A[i]]++;
            arrB[B[i]]++;
        }
        for (int i = 0; i < 6; ++i)
        {
            int j;
            int tim = find(arrA, 7, size);
            //if (tim == -1) return 0;
            for (j = 0; j < size; ++j)
            {
                if (A[j] != tim && B[j] != tim)
                {
                    break;
                }
            }
            if (j == size)
            {
                leftMin = size - arrA[tim];
                break;
            }
            arrA[tim] = 0;
        }
        for (int i = 0; i < 6; ++i)
        {
            int j;
            int tim = find(arrB, 7, size);
           // if (tim == -1) return 0;
            for (j = 0; j < size; ++j)
            {
                if (B[j] != tim && A[j] != tim)
                {
                    break;
                }
            }
            if (j == size)
            {
                rightMin = size - arrB[tim];
                break;
            }
            arrB[tim] = 0;
        }
        if (leftMin != -1 && rightMin != -1)
            return leftMin < rightMin ? leftMin : rightMin;
        else if (leftMin == -1) return rightMin;
        else return leftMin;
    }
};

剑指offer：二叉树的下一个节点

class Solution {
public:
    TreeLinkNode* GetNext(TreeLinkNode* root)
    {
        if(root == NULL) return NULL;
        TreeLinkNode* r = root;
        if(root->right != NULL)
        {
            TreeLinkNode* temp = root->right;
            while(temp->left != NULL)
            {
                temp = temp->left;
            }
            return temp;
        }
        else
        {
            while(root->next != NULL)
            {
                 if(root->next->left == root) return root->next;
                 else
                 {
                     root = root->next;
                 }
            }
            return NULL;
        }
        
    }
};

剑指offer：字符流中第一个不重复的字符

class Solution
{
public:
    int arr[256];
    int index;
    Solution()
    {
        index = 0;
        for(int i = 0;i<256;++i) arr[i] = -1;
    }
  //Insert one char from stringstream
    void Insert(char ch)
    {
        int t = (int)ch;
         if(arr[t] == -1) arr[t] = index;
         else arr[t] = -2;
        ++index;
    }
  //return the first appearence once char in current stringstream
    char FirstAppearingOnce()
    {
        int MinIndex = 0x7fffffff;
        char result = '#';
        for(int i=0;i<256;++i)
        {
            if(arr[i] >= 0 && MinIndex > arr[i])
            {
                MinIndex = arr[i];
                result = (char)i;
            }
        }
        return result;
        
    }

};

剑指offer：链表中环的入口节点

class Solution {
public:
    ListNode* MeetingNode(ListNode* head)
    {
        if(head == NULL) return NULL;
        ListNode* slow = head->next;
        if(slow == NULL) return NULL;
        ListNode* fast = slow->next;
        while(slow!=NULL && fast!=NULL)
        {
            if(slow == fast) return fast;
            slow = slow->next;
            fast = fast->next;
            if(fast->next != NULL)
            {
                fast = fast->next;
            }
            else return NULL;
        }
            
        return NULL;
    }
    ListNode* EntryNodeOfLoop(ListNode* head)
    {
        if(head == NULL) return NULL;
        ListNode* current = MeetingNode(head);
        if(current == NULL) return NULL;
        int count = 1;
        ListNode* tim = current->next;
        while(tim != current)
        {
            ++count;
            tim = tim->next;
        }
        
        ListNode* fast = head;
        ListNode* slow = head;
        while(count>0)
        {
            fast = fast->next;
            --count;
        }
        while(slow != fast)
        {
            slow=slow->next;
            fast=fast->next;
        }
        return fast;
    }
};

leetcode：997. 找到小镇的法官

class Solution {
public:
    int findJudge(int N, vector>& trust) 
    {
        int re = 0;
        int result = -1;
        if(N == 1) return 1;
        if(trust.empty()) return -1;
        int *arr = new int[N+1];
        for(int i=0;i<=N;++i) arr[i] = 0;
        mapm;
        map::iterator it;
        int size = trust.size();
        for(int i=0;i

leetcode：41. 缺失的第一个正数

class Solution {
public:
    int firstMissingPositive(vector& nums) 
    {
        int size = nums.size();
        if(size == 0) return 1;
        for(int i=0;i0 && (nums[i] <= size) && nums[nums[i]-1] != nums[i]) 
                {
                    swap(nums[i],nums[nums[i]-1]);
                }
                else ++i;
            }
            else ++i;
        }
        for(int i=0;i

剑指offer：二叉搜索树的第k个节点

class Solution {
public:
    vector v;
    void fun(TreeNode* root)
    {
        if(root == NULL) return;
        fun(root->left);
        v.push_back(root);
        fun(root->right);
    }
    TreeNode* KthNode(TreeNode* root, int k)
    {
        if(k == 0 || root == NULL) return NULL;
        fun(root);
        if(k<=v.size()) return v[k-1];
        else return NULL;
    }
}  

剑指offer：把二叉树打印成多行

class Solution {
public:
        vector > Print(TreeNode* root) 
        {
            queue q;
            vector> result;
            int current = 1;
            int nexth = 0;
            if(root == NULL) return result;
            TreeNode* p = root;
            q.push(p);
            while(!q.empty())
            {
                
                vector v;
                while(current)
                {
                    TreeNode* t = q.front();
                    v.push_back(t->val);
                    if(t->left != NULL) 
                    {
                        q.push(t->left);
                        ++nexth;
                    }
                    if(t->right != NULL)
                    {
                        q.push(t->right);
                        ++nexth;
                    }
                    q.pop();
                    --current;
                }
                current = nexth;
                nexth = 0;
                result.push_back(v);
                v.clear();
            }
            return result;
        }
    
};

剑指offer：按之子型打印二叉树

class Solution {
public:
    vector > Print(TreeNode* root)
    {
        vector> result;
        if(root == NULL) return result;
        stack qi;
        stack ou;
        qi.push(root);
        int hang = 1;
        vector v;
        while(!qi.empty() || !ou.empty())
        {
            if(hang&1)
            {
                v.clear();
                while(!qi.empty())
                {
                    TreeNode* tim = qi.top();
                    v.push_back(tim->val);
                    if(tim->left != NULL) ou.push(tim->left);
                    if(tim->right != NULL) ou.push(tim->right);
                    qi.pop();
                }
                ++hang;
                result.push_back(v);
            }
            else
            {
                v.clear();
                while(!ou.empty())
                {
                    TreeNode* tim = ou.top();
                    v.push_back(tim->val);
                    if(tim->right != NULL) qi.push(tim->right);
                    if(tim->left != NULL) qi.push(tim->left);
                    ou.pop();
                }
                ++hang;
                result.push_back(v);
            }
        }
        return result;
    }
      
};

剑指offer：对称的二叉树

class Solution {
public:
    bool isSymmetrical(TreeNode* root1,TreeNode* root2)
    {
        if(root1==NULL && root2==NULL) return true;
        if(root1==NULL || root2==NULL) return false;
        if(root1->val == root2->val)
        {
            return isSymmetrical(root1->left,root2->right) && isSymmetrical(root1->right,root2->left);
        }
        return false;
    }
    bool isSymmetrical(TreeNode* root)
    {
        return isSymmetrical(root,root);
    }
 
};

leetcode：

26. 删除排序数组中的重复项

class Solution {
public:
    int removeDuplicates(vector& nums) 
    {
        int size = nums.size();
        if(size <= 1) return size;
        int fast=1,slow=0;
        while(fast

leetcode：

33. 搜索旋转排序数组

class Solution {
public:
    int fun(vector& nums, int target,int left, int right)
    {
        int size = nums.size();
        if(size == 0) return -1;
        if(left < 0 || right >=size) return -1;
        int begin = left;
        int end = right;
        int midan;
        while(begin <= end)
        {
            midan = (begin + end)/2;
            if(nums[midan] == target) return midan;
            if(nums[midan] < target) begin = midan+1;
            else if(nums[midan] > target) end = midan-1;
        }
        return -1;
    }
    int search(vector& nums, int target) 
    {
        int size = nums.size();
        if(size == 0) return -1;
        else if(size == 1) return nums[0] == target?0:-1;
       
        
        if(nums[0] < nums[size-1])
        {
            return fun(nums,target,0,size-1);
        }
        else
        {
            int midan;
            int begin = 0,end = size-1;
            while(begin < end-1)
            {
                midan = (begin+end)/2;
                if(nums[midan] > nums[end]) begin = midan;
                else if(nums[midan] < nums[begin]) end = midan;
            }
            int min = end;
            int result = fun(nums,target,0,min-1);
            if(result != -1) return result;
            return fun(nums,target,min,size-1);
        }
        
    }
};

leetcode：80. 删除排序数组中的重复项 II

class Solution {
public:
    int removeDuplicates(vector& nums) 
    { 
        int size = nums.size();
        if(size <= 2) return size;
        int fast=1,slow=0;
        int current = nums[slow];
        int count = 1;
        while(fast

leetcode：153. 寻找旋转排序数组中的最小值

class Solution {
public:
    int findMin(vector& nums) 
    {
        int size = nums.size();
        if(size == 0) return -1;
        if(size == 1) return nums[0];
        if(nums[0] < nums[size-1]) return nums[0];
        
        int begin = 0;int end = size-1;
        int midan;
        while(begin < end-1)
        {
            midan = (begin+end)/2;
            if(nums[midan] > nums[begin]) begin = midan;
            else if(nums[midan] < nums[end]) end = midan;
        }
        return nums[end];
    }
};

算法第四版：二叉搜索树增删改查的实现

class Node
{
public:
	int key;
	char value;
	Node* left;
	Node* right;
	int N;
	Node()
	{

	}
	Node(int key, char value, int N)
	{
		this->key = key;
		this->value = value;
		this->N = N;
	}
};
class BST:Node
{
	Node* root;
public:
	BST()
	{
		root = NULL;
	}
	int size(Node* r)
	{
		if (r == NULL) return 0;
		return r->N;
	}
	char get(int key);
	char get(Node* x, int key);
	void insert(int key, char value);
	Node* insert(Node* x, int key, int value);
	int min();
	Node* min(Node* x);
	int select(int k);
	Node* select(Node* x, int k);
	int rank(int key);
	int rank(int key, Node* x);
	void deleteMin();
	Node* deleteMin(Node* x);
	void delete_BST(int key);
	Node* delete_BST(Node* x, int key);

};
char BST::get(int key)
{
	char result = '\0';
	if (root == NULL) return result;
	return get(root, key);
}
char BST::get(Node* x, int key)
{
	char result = '\0';
	if (x == NULL) return result;
	if (x->key == key) return x->value;
	if (x->key < key) return get(x->right, key);
	if (x->key > key) return get(x->left, key);
}
void BST::insert(int key, char value)
{
	root = insert(root, key, value);
}

Node* BST::insert(Node* x, int key, int value)
{
	if (x == NULL)
	{
		Node* NewNode = new Node(key,value,1);
		return NewNode;
	}
	if (x->key < key) x->right = insert(x->right, key, value);
	else if (x->key > key) x->left = insert(x->left, key, value);
	else if (x->key == key) x->value = value;

	x->N = size(x->left) + size(x->right) + 1;
	return x;
}

int BST::min()
{
	Node* result = min(root);
	if (root == NULL) return 0;
	return result->key;
	
}
Node* BST::min(Node* x)
{
	if (x == NULL) return NULL;
	if (x->left != NULL) return min(x->left);
	return x;
}

int BST::select(int k)
{
	Node* result = select(root, k);
	if (result == NULL) return 0;
	return result->key;
}
Node* BST::select(Node* x, int k)
{
	if (x == NULL) return NULL;
	int t = 0;
	if (x->left != NULL) t = size(x->left);
	if (t < k) return select(x->right, k - 1 - t);
	if (t == k) return x;
	if (t > k) return select(x->left, k);
}

void BST::deleteMin()
{
	if (root == NULL) return;
	Node* result = deleteMin(root);
	//delete result;
	//result = NULL;
}
/*
Node* BST::deleteMin(Node* x)
{
	if (x == NULL) return NULL;
	Node* dMin = min(x);
	if (dMin == NULL) return NULL;
	*dMin = *(dMin->right);
	return dMin;
}*/
Node* BST::deleteMin(Node* x)
{
	if (x == NULL) return NULL;
	if (x->left == NULL) return x->right;
	x->left = deleteMin(x->left);
	x->N = size(x->left) + size(x->right) + 1;
	return x;
}

int BST::rank(int key)
{
	return rank(key, root);
}
int BST::rank(int key, Node* x)
{
	if (x == NULL) return 0;
	int t = x->left == NULL ? 0 : x->left->N;
	if (x->key > key) return rank(key, x->left);
	if (x->key == key) return t;
	if (x->key < key) return rank(key, x->right) + t + 1;
}

void BST::delete_BST(int key)
{
	delete_BST(root, key);
}
Node* BST::delete_BST(Node* x, int key)
{
	if (x == NULL) return NULL;
	if (key == x->key)
	{
		if (x->right == NULL) return x->left;
		if (x->left == NULL) return x->right;
		
		Node* p = x;
		x = min(x->right);
		if (x == NULL) return x->left;
		x->left = p->left;
		x->right = deleteMin(p->right);
		delete p;
		p = NULL;
	}
	else if (key < x->key) x->left = delete_BST(x->left, key);
	else if (key > x->key) x->right = delete_BST(x->right, key);
	x->N = size(x->left) + size(x->right) + 1;
	return x;
}

leetcode:154. 寻找旋转排序数组中的最小值 II

class Solution {
public:
    int sort(vector& nums)
    {
        int size = nums.size();
        int result = nums[0];
        for(int i = 1;i nums[i] ? nums[i] : result;
        return result;
    }
    int findMin(vector& nums) 
    {
        int result;
        int size = nums.size();
        if(size == 1) return nums[0];
        int left = 0;
        int right = size-1;
        if(nums[0] < nums[right]) return nums[0];
        int midan;
        while(left < right-1)
        {
            midan = (left+right)>>1;
            if(nums[midan] == nums[left] && nums[midan] == nums[right]) 
            {
                result = sort(nums);
                return result;
            }
            if(nums[midan] < nums[left]) right = midan;
            else if(nums[midan] > nums[right]) left = midan;
        }
        return nums[right];
    }
};

leetcode:11. 盛最多水的容器

class Solution {
public:
    int maxArea(vector& height) 
    {
        int size = height.size();
        if(size <= 1) return 0;
        int left = 0;
        int right = size-1;
        int result = 0;
        while(left ((right - left)*tim) ? result : ((right - left)*tim);
            if(height[left] < height[right]) ++left;
            else --right;
        }
        return result;
    }
};

leetcode：题目号：200 575 695 268  674 135 代码略

leetcode:560 209 112 113 108

左神书题：最大值减去最小值小于等于num的子数组数量

leetcode:795

class Solution {
public:
    int numSubarrayBoundedMax(vector& arr, int max)
    {
        int ret = 0;
        int temp = 0;
        int size = arr.size();
        for(int i=0;i& arr, int L, int R) 
    {
        // 最大元素满足大于等于L小于等于R的子数组个数 = 最大元素小于等于R的子数组个数 - 最大元素小于L的子数组个数
        int size = arr.size();
        if(size == 0) return 0;
        return numSubarrayBoundedMax(arr,R)-numSubarrayBoundedMax(arr,L-1);
    }
};

leetcode: 98 35 104 110 111 102

 

leetcode:96不同的二叉搜索树

/*结题思路：假设n个节点存在二叉排序树的个数是G(n)，1为根节点，2为根节点，...，n为根节点，当1为根节点时，其左子树节点个数为0，右子树节点个数为n-1，同理当2为根节点时，其左子树节点个数为1，右子树节点为n-2，所以可得G(n) = G(0)*G(n-1)+G(1)*(n-2)+...+G(n-1)*G(0)*/

class Solution {
    public int numTrees(int n) 
    {
        if(n<2) return 1;
        int[] arr = new int [n+1];
        arr[0] = 1;
        arr[1] = 1;
        
        for(int i=2;i

leetcode:100，94，230，144，116，117，111 ,129  130  199   96  137 141 142  287 101 222  235 236 404
在两个长度相等的排序数组中找到上中位数
84 85 221 513  542  718  581  27  283  485  1004 

1 15 34 42 53 121 122 62 63
75 88 167 653 344  345 215
342 231 191 338  349 350 

242  438 290  205 525 

451 347  49  快速幂 1002 56 204