LeetCode刷算法题-简单难度（四）题号235-383

235. 二叉搜索树的最近公共祖先
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(p->val < root->val && q->val < root->val)
            return lowestCommonAncestor(root->left,p,q);
        if(p->val > root->val && q->val > root->val)
            return lowestCommonAncestor(root->right,p,q);
        return root;
    }
};

237. 删除链表中的节点
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        if(node == NULL || node->next == NULL)
            return;
        ListNode* temp = node->next;
        node->val = temp->val;
        node->next = temp->next;
        delete temp;
    }
};

242. 有效的字母异位词
class Solution {
public:
    bool isAnagram(string s, string t) {
        if(s.size() != t.size())
            return false;
        char sArr[26] = {0};
        char tArr[26] = {0};
        for(int i = 0; i < s.size(); ++i){
            sArr[s[i]-'a']++;
            tArr[t[i]-'a']++;
        }
        for(int i = 0; i < 26; ++i){
            if(sArr[i]!=tArr[i])
                return false;
        }
        return true;
    }
};

257. 二叉树的所有路径
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if(root==NULL)
            return res;
        if(root->left == NULL && root->right == NULL)
            res.push_back(to_string(root->val));
        vector<string> leftS = binaryTreePaths(root->left);
        for(int i = 0; i < leftS.size(); ++i){
            res.push_back(to_string(root->val)+"->"+leftS[i]);
        }
        vector<string> rightS = binaryTreePaths(root->right);
        for(int i = 0; i < rightS.size(); ++i){
            res.push_back(to_string(root->val)+"->"+rightS[i]);
        }
        return res;
    }
};

258. 各位相加
class Solution {
public:
    int addDigits(int num) {
        return (num-1)%9+1;
    }
};

263. 丑数
class Solution {
public:
    bool isUgly(int num) {
        if(num == 0)
            return false;
        while(num%2==0)
            num /= 2;
        while(num%3==0)
            num /= 3;
        while(num%5==0)
            num /= 5;
        if(num == 1)
            return true;
        else
            return false;
    }
};

268.缺失数字
class Solution {
public:
    int missingNumber(vector<int>& nums) {
        if(nums.size() == 1)
            return 1 - nums[0];
        sort(nums.begin(), nums.end());
        if(nums[0]!=0)
            return 0;
        for(int i = 1; i < nums.size(); ++i){
            if(nums[i]-nums[i-1] != 1)
                return nums[i-1]+1;
        }
        return nums.size();
    }
};

278.第一个错误的版本
// The API isBadVersion is defined for you.
// bool isBadVersion(int version);

class Solution {
public:
    int firstBadVersion(int n) {
        int low = 1;
        int high = n;
        while(low <= high){
            int mid = low+(high-low)/2;
            if(isBadVersion(mid)){
                high = mid - 1;
            }else{
                low = mid + 1;
            }
        }
        return low;
    }
};

283. 移动零
class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        if(nums.size() <= 1)
            return;
        int k = 0;
        for(int i = 0; i < nums.size(); ++i){
            if(nums[i]!=0){
                if(k!=i){
                    swap(nums[k],nums[i]);
                }
                k++;
            }
        }
    }
};

290. 单词规律
class Solution {
public:
    bool wordPattern(string pattern, string str) {
        map<char, string> mm;
        istringstream in(str);
        string word;
        int i = 0;
        while(in>>word){
            if(mm.find(pattern[i])!=mm.end()){
                if(mm[pattern[i]] != word)
                    return false;
            }else{
                for(auto &p : mm){
                if(p.second == word)
                    return false;
                }
                mm[pattern[i]] = word;            
            }
            ++i;
        }
        if(i == pattern.size())
            return true;
        else
            return false;
    }
};

292. Nim 游戏
class Solution {
public:
    bool canWinNim(int n) {
        return n%4==0?false:true;
    }
};

303. 区域和检索 - 数组不可变
class NumArray {
public:
    NumArray(vector<int>& nums) {
        arr = nums;
    }
    
    int sumRange(int i, int j) {
        int sum = 0;
        for(int k = i; k <= j; ++k){
            sum += arr[k];
        }
        return sum;
    }
private:
    vector<int> arr;
};

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray* obj = new NumArray(nums);
 * int param_1 = obj->sumRange(i,j);
 */

326. 3的幂
class Solution {
public:
    bool isPowerOfThree(int n) {
        if(n<=0)
            return false;
        int k = log(0x7fffffff)/log(3);
        int b = pow(3,k);
        return b%n == 0;
    }
};

342. 4的幂
class Solution {
public:
    bool isPowerOfFour(int num) {
        long n = (long)num;
        return ((n&n-1) == 0) && (n > 0) && ((n&0x55555555) != 0);
    }
};

344. 反转字符串
class Solution {
public:
    void reverseString(vector<char>& s) {
        if(s.empty())
            return;
        int left = 0, right = s.size()-1;
        while(left < right){
            swap(s[left++],s[right--]);
        }
    }
};

345. 反转字符串中的元音字母
class Solution {
public:
    string reverseVowels(string s) {
        if(s.empty())
            return s;
        string vowel = "aeiouAEIOU";
        int left = 0, right = s.length()-1;
        while(left < right){
            if(vowel.find(s[left])== string::npos)
                left++;
            else if(vowel.find(s[right])== string::npos)
                right--;
            else {
                swap(s[left++],s[right--]);
            }
        }
        return s;
    }
};

349. 两个数组的交集
class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        set<int> set1(nums1.begin(),nums1.end());
        set<int> res;
        for(int i = 0; i < nums2.size(); ++i){
            if(set1.find(nums2[i])!= set1.end())
                res.insert(nums2[i]);
        }
        return vector<int>(res.begin(), res.end());
    }
};

350. 两个数组的交集 II
class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        map<int,int> map1;
        vector<int> res;
        for(int i = 0; i < nums1.size(); ++i){
            map1[nums1[i]]++;
        }
        for(int i = 0; i < nums2.size(); ++i){
            if(map1[nums2[i]] > 0){
                res.push_back(nums2[i]);
                map1[nums2[i]]--;
            }
        }
        return res;
    }
};

367. 有效的完全平方数
class Solution {
public:
    bool isPerfectSquare(int num) {
        long left = 1, right = (long)num;
        while(left <= right){
            long mid = left + (right-left)/2;
            if(mid*mid == num)
                return true;
            else if(mid*mid > num)
                right = mid -1;
            else 
                left = mid + 1;
        }
        return false;
    }
};

371. 两整数之和
class Solution {
public:
    int getSum(int a, int b) {
        while(b){
            auto c = ((unsigned int)a&b) << 1;
            a = a ^ b;
            b = c;
        }
        return a;
    }
};

374.猜数字大小
/** 
 * Forward declaration of guess API.
 * @param  num   your guess
 * @return 	     -1 if num is lower than the guess number
 *			      1 if num is higher than the guess number
 *               otherwise return 0
 * int guess(int num);
 */

class Solution {
public:
    int guessNumber(int n) {
        int left = 1, right = n;
        while(left < right){
            int mid = left + (right-left)/2;
            if(guess(mid) == 1)
                left = mid + 1;
            else if(guess(mid) == -1)
                right = mid;
            else 
                return mid;
        }
        return n;
    }
};

383. 赎金信
class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        map<char,int> m1;
        for(int i = 0; i < magazine.length(); ++i){
            m1[magazine[i]]++;
        }
        for(int i = 0; i < ransomNote.length(); ++i){
            m1[ransomNote[i]]--;
            if(m1[ransomNote[i]] < 0)
                return false;
        }
        return true;
    }
};