Leetcode 子序列 53 392 115 583

子序列：

53. Maximum Subarray

class Solution {
public:
    int maxSubArray(vector& nums) {
        if(nums.size() == 1) return nums[0];
        vector dp(nums.size(), 0);
        dp[0] = nums[0];
        int res = nums[0];

        for(int i=1; i

392. Is Subsequence

1.双指针

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int i=0, j=0;
        while(i

2.动态规划

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if(s.size() > t.size()) return false;
        vector> dp(s.size()+1, vector (t.size()+1, 0));

        for(int i=1; i<=s.size(); i++){
            for(int j=1; j<=t.size(); j++){
                if(s[i-1] == t[j-1]){
                    dp[i][j] = dp[i-1][j-1]+1;
                }else{
                    dp[i][j] = dp[i][j-1];
                }
            }
        }

        if(dp[s.size()][t.size()] == s.size()) return true;
        return false;
    }
};

115. Distinct Subsequences

class Solution {
public:
    int numDistinct(string s, string t) {
        vector> dp(s.size()+1, vector (t.size()+1, 0));
        for(int i=0; i<=s.size(); i++){
            dp[i][0] = 1;
        }

        for(int i=1; i<=s.size(); i++){
            for(int j=1; j<=t.size(); j++){
                if(s[i-1] == t[j-1]){
                    dp[i][j] = dp[i-1][j-1]+dp[i-1][j];
                }else{
                    dp[i][j] = dp[i-1][j];
                }
            }
        }

        return dp[s.size()][t.size()];
    }
};

初始化：

这里感觉对我来说有点难以理解

dp[i][0] 这里我的理解是，以下标i-1位结尾的s和空集能够相同的子序列的个数，那么肯定只有一个，就是把s里的全部删掉

dp[0][j]s是空数组所以不可能和t一样（除了t是空数组的时候，所以dp[0][0]为1）

注意：

这里不能直接用int会报错

583. Delete Operation for Two Strings

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector> dp(word1.size()+1, vector (word2.size()+1, 0));
        for(int i=0; i<=word1.size(); i++) dp[i][0] = i;
        for(int j=0; j<=word2.size(); j++) dp[0][j] = j;

        for(int i=1; i<=word1.size(); i++){
            for(int j=1; j<=word2.size(); j++){
                if(word1[i-1] == word2[j-1]){
                    dp[i][j] = dp[i-1][j-1];
                }else{
                    dp[i][j] = min(dp[i-1][j]+1, dp[i][j-1]+1);
                }
            }
        }

        return dp[word1.size()][word2.size()];

    }
};

二叉树：

102. Binary Tree Level Order Traversal

class Solution {
public:
    vector> levelOrder(TreeNode* root) {
        queue que;
        if(root != NULL) que.push(root);
        vector> res;
        while(!que.empty()){
            int size = que.size();
            vector vec;
            for(int i=0; ival);
                que.pop();
                if(cur->left) que.push(cur->left);
                if(cur->right) que.push(cur->right);
            }
            res.push_back(vec);
        }
        return res;
    }
};

层序遍历

107. Binary Tree Level Order Traversal II

class Solution {
public:
    vector> levelOrderBottom(TreeNode* root) {
        queue que;
        if(root != NULL) que.push(root);
        vector> res;

        while(!que.empty()){
            int size = que.size();
            vector vec;
            for(int i=0; ival);
                if(cur->left) que.push(cur->left);
                if(cur->right) que.push(cur->right);
            }

            res.push_back(vec);
        }

        reverse(res.begin(), res.end());
        return res;
    }
};