Leetcode-W1，2

week 1

双指针(题号：167)：https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2. 

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int left=0; int right=numbers.size() - 1;
        while(right > left){
            if(numbers[left] + numbers[right] == target)
                return {left+1,right+1};
            else if(numbers[left] + numbers[right] > target)
                right--;
            else left++;
        }
        return {};
    }
};

View Code

排序：

快速选择、堆排序（题号：215）：https://leetcode.com/problems/kth-largest-element-in-an-array/description/

Input: [3,2,3,1,2,4,5,5,6] and k = 4
Output: 4

 1.quickSelect:

class Solution {
    public void Sort(int[] t, int left,int right) {
		if(left>right) return ;
		int i = left;
		int j = right;
		int temp = t[left];
		while(i=temp && i

class Solution {
    public int findKthLargest(int[] nums, int k) {
        int i = 0;
        int j = nums.length - 1;
        int partitionIdx;
        
        while(i <= j){
            partitionIdx = partition(nums, i, j);
            
            if((k - 1) == partitionIdx){
                return nums[partitionIdx];
            }
            else if((k -1) < partitionIdx){
                j = partitionIdx - 1;
            }
            else{
                i = partitionIdx + 1;
            }
        }
        return 0;
    }
    
    public int partition(int[] nums, int start, int end){
        if(start == end){
            return start;
        }
        int pivot = nums[start];
        while(start < end){
            while(start < end && nums[end] <= pivot){
                end--;
            }
            nums[start] = nums[end];
            
            while(start < end && nums[start] >= pivot){
                start++;
            }
            nums[end] = nums[start];
        }
        nums[start] = pivot;
        return start;
    }
}

View Code

2.Heap(priorityQueue)

class Solution {
    public int findKthLargest(int[] nums, int k) {
        PriorityQueue pq = new PriorityQueue<>();
        for(int val : nums) {
            pq.offer(val);
            
            if(pq.size() > k) {
                pq.poll();
            }
        }
        return pq.peek();
    }
}

View Code

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
    // min heap
    priority_queue< int, vector<int>, greater<int> > pq;
    
    // add first k elements to heap, since min heap, min element will be at top
    for (int i=0; ii) {
        pq.push(nums[i]);
    }
    
    // for rest of elements...
    // ... if it is greater than heap's top...
    // ... pop from heap and push num to heap
    // thereby maintaining heap size to be (k)
    
    // one by one, all numbers in heap become the (k) largest numbers in array
    
    for (int i=k; ii) {
        if (nums[i] > pq.top()) {
            pq.pop();
            pq.push(nums[i]);
        }
    }
    
    // at the end, the number at top will be less than all numbers in heap
    // that is, it will be kth largest
    
    return pq.top();
    }
};

View Code

 桶排序（题号：347）：https://leetcode.com/problems/top-k-frequent-elements/description/

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int,int> um;
        // map number frequency
        for (int c: nums) um[c]++;
        priority_queue< pair<int,int> > pq; 
        // max heap by frequency，默认大顶堆
        for (auto i: um) 
            pq.push(make_pair(i.second,i.first));
        vector<int> ans;
        // for each element of heap,
        // add that num to answer
        while (k--) {
            auto x = pq.top();
            pq.pop();

            ans.push_back(x.second);
            }
        return ans;
    }

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 荷兰国旗问题（题号：75）：https://leetcode.com/problems/sort-colors/description/

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

void sortColors(vector<int>& nums) {
        int end_zero=0;
        int end_one=0;
        int end_two=0;
        int n=nums.size();
        for(int i=0;i){
            switch(nums[i]){
                case 0:
                    swap(nums[i],nums[end_zero]);
                    end_zero++;
                    if(nums[i]!=1){
                        end_one++;
                        end_two++;
                        break;
                    }
                case 1:
                    swap(nums[i],nums[end_one]);
                    end_one++;
                    end_two++;
                    break;
                case 2:
                    swap(nums[i],nums[end_two]);
                    end_two++;
                    break;
                i++;
            }
        }

View Code

贪心（题号：455）：https://leetcode.com/problems/assign-cookies/description/

Input: [1,2], [1,2,3]
Output: 2

int findContentChildren(vector<int>& g, vector<int>& s) {
        int contentLevel = 0;
        
        sort(g.begin(), g.end(), greater<int>()); //sort in reverse for easier popping
        sort(s.begin(), s.end(), greater<int>());
        
        while(g.size() > 0 && s.size() > 0){//just a greedy algorithm where both are consumed
            if(g[g.size()-1] > s[s.size()-1])//when satisfaction is greater than greed
                s.pop_back();
            else{
                s.pop_back();
                g.pop_back();
                contentLevel++;
            }
                
        }
        return contentLevel;
    }

View Code

二分（题号：69）：https://leetcode.com/problems/sqrtx/description/ 

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842...

public int mySqrt(int x) {
    if (x <= 0) return 0;
    long res = x;
    while (res * res > x) res = (res + x/res) / 2;
    return (int)res;
    }

分治（题号：241）：https://leetcode.com/problems/different-ways-to-add-parentheses/description/

Input: "2*3-4*5"
Output: [-34, -14, -10, -10, 10]
Explanation:
(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10

class Solution {
  private Map> resultMap = new HashMap<>();
  public List diffWaysToCompute(String input) {
    if (resultMap.containsKey(input)) return resultMap.get(input);
      
    List ans = new ArrayList<>();
    
    for (int i = 0; i < input.length(); i ++) {
        
      char op = input.charAt(i);
      if (op == '+' || op == '-' || op == '*') {
          
        String left = input.substring(0, i);
        String right = input.substring(i + 1);
        
        List l = diffWaysToCompute(left);
        List r = diffWaysToCompute(right);
        
        switch (op) {
          case '+': {
            for (int a : l) 
              for (int b : r) 
                ans.add(a + b);
            break;
          }
            
          case '-': {
            for (int a : l) 
              for (int b : r) 
                ans.add(a - b);
            break;
          }
            
          case '*': {
            for (int a : l) 
              for (int b : r) 
                ans.add(a * b);
            break;
          }
        }
      }
    }
    
    if (ans.size() == 0) ans.add(Integer.valueOf(input));
    
    resultMap.put(input, ans);
    return ans;
  }
}

链表（题号：160）：https://leetcode.com/problems/intersection-of-two-linked-lists/description/

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if(headA==null || headB==null){return null;}
        Set set=new HashSet();
        while(headA!=null){
            set.add(headA);
            headA=headA.next;
            }
        while(headB!=null){
            if(set.contains(headB)){return headB;}
            headB=headB.next;
            }
        return null;
    }

哈希表（题号：1）：https://leetcode.com/problems/two-sum/description/

Given nums = [2, 7, 11, 15], target = 9

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1]

public int[] twoSum(int[] nums, int target) {
        
        Map index = new HashMap<>();
        int len = nums.length;
        int pair[] = new int[2];
        
        for(int i = 0 ; i < len ; i++){
            
            if(null != index.get((target-nums[i]))){
                
                pair[1] = i;
                pair[0] = index.get((target-nums[i]));
                return pair;
            }
            else{
                index.put(nums[i],i);
            }
        }
        return pair;
    }

字符串（题号：242）：https://leetcode.com/problems/valid-anagram/description/

Input: s = "anagram", t = "nagaram"
Output: true

public boolean isAnagram(String s, String t) {
        int[] cntS = new int[256];
        int[] cntT = new int[256];
        for (char c : s.toCharArray()) {
            cntS[c]++;
        }
        for (char c : t.toCharArray()) {
            cntT[c]++;
        }
        for (int i = 0; i < 256; i++) {
            if (cntS[i] != cntT[i]) {
                return false;
            }
        }
        return true;
    }

栈和队列（题号：232）：https://leetcode.com/problems/implement-queue-using-stacks/description/

public class MyQueue {
    Stack stack;

    /**
     * Initialize your data structure here.
     */
    public MyQueue() {
        stack = new Stack<>();
    }

    /**
     * Push element x to the back of queue.
     */
    public void push(int x) {
        stack.add(x);

    }

    /**
     * Removes the element from in front of queue and returns that element.
     */
    public int pop() {
        if (!empty()) {
            Iterator it = stack.iterator();
            while (it.hasNext()) {
                Integer value = it.next();
                if (!it.hasNext()) ;
                it.remove();
                return value;
            }
        }

        return Integer.parseInt(null);
    }

    /**
     * Get the front element.
     */
    public int peek() {
        if (!empty()) {
            Iterator it = stack.iterator();
            while (it.hasNext()) {
                Integer value = it.next();
                if (!it.hasNext()) ;
                return value;
            }
        }

        return Integer.parseInt(null);
    }

    /**
     * Returns whether the queue is empty.
     */
    public boolean empty() {
        return stack.empty();
    }
}

 week 2

字符串：409. Longest Palindrome（回文） https://leetcode.com/problems/longest-palindrome/description/

Input: "abccccdd"

Output: 7

Explanation: One longest palindrome that can be built is "dccaccd", whose length is 7

public int longestPalindrome(String s) {
        if(s.length()==0) return 0;
        
        int[] chars=new int[128];
        for(char c:s.toCharArray()) chars[c]++;
        
        int cnt=0;
        for(int i=0;i<128;i++) cnt+=chars[i]/2;
        
        cnt=cnt*2;
        return cnt==s.length()?cnt:cnt+1;//如果不是全部用光偶数个，那一定有一个可以放中间
}

 数组和矩阵：378. Kth Smallest Element in a Sorted Matrix https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/

public int kthSmallest(int[][] matrix, int k) {
        if(matrix.length * matrix[0].length < k) return -1;
        
        PriorityQueue q=new PriorityQueue(k,Comparator.reverseOrder());
        
        int len=0;
        for(int i=0;i matrix[i][j]) {
                    q.poll();
                    q.offer(matrix[i][j]);
                }
            }
        }
        return q.peek();
    }

 位运算：260. Single Number III https://leetcode.com/problems/single-number-iii/description/

 Input:  [1,2,1,3,2,5]
Output: [3,5]

public int[] singleNumber(int[] items) {
        int[] result = new int[2];
        int xor = 0;
        for(int i : items){
            xor ^= i;
        }
        xor &= ~(xor - 1);
        for(int i : items){
            if((i&xor) != 0){
                result[0] ^= i;
            }
            else{
                result[1] ^= i;
            }
        }
        return result;
    }

 数学

进制转换：转化为7进制 https://leetcode.com/problems/base-7/description/

public String convertToBase7(int num) {
        String result = "";
        int q = num;
        int r;
        boolean isNeg = false;
        
        if(q < 0) {
            q = Math.abs(q);
            isNeg = true;
        }
        
        do {
            r = q % 7;
            q /= 7;
            result += r;
        } while (q != 0);
        
        if (isNeg)
            return "-" + reverse(result);
        else
            return reverse(result);
    }
    
    public String reverse(String input){
        byte [] bytes = input.getBytes(); 
  
        byte [] result = new byte [bytes.length]; 
        
        for (int i = 0; i < bytes.length; i++) 
            result[i] = bytes[bytes.length-i-1]; 
  
        return new String(result);
    }

相遇问题：462. Minimum Moves to Equal Array Elements II https://leetcode.com/problems/minimum-moves-to-equal-array-elements-ii/description/

Input:[1,2,3]

Output:2

Explanation:Only two moves are needed (remember each move increments or decrements one element):[1,2,3]  =>  [2,2,3]  =>  [2,2,2]

public int minMoves2(int[] A) {
    int sum = 0, median = quickselect(A, A.length/2+1, 0, A.length-1);
    for (int i=0;i pivot) r--;
        if (l>=r) break;
        swap(A, l++, r--);
    }
    if (l-start+1 > k) return quickselect(A, k, start, l-1);
    if (l-start+1 == k && l==r) return A[l];
    return quickselect(A, k-r+start-1, r+1, end);
}

public void swap(int[] A, int i, int j) {
    int temp = A[i];
    A[i] = A[j];
    A[j] = temp;
}

 多数投票问题：169. Majority Element https://leetcode.com/problems/majority-element/description/

public int majorityElement(int[] num) {
//      O(n) time O(1) space
        int major=num[0], count = 1;
        for(int i=1; i

转载于:https://www.cnblogs.com/zhangzx/p/11391107.html