第一题、最长公共子序列 力扣题目链接

class Solution {
public:int longestCommonSubsequence(string text1, string text2) {vector<vector<int>> dp(text1.size()+1, vector<int>(text2.size()+1, 0));for(int i=1; i < text1.size()+1; i++){for(int j=1; j < text2.size()+1; j++){if(text1[i-1] == text2[j-1]){dp[i][j] = dp[i-1][j-1] + 1;}else{dp[i][j] = max(dp[i-1][j], dp[i][j-1]);}}}return dp[text1.size()][text2.size()];}
};

第二题、不想交的线 力扣题目链接

跟上题是一毛一样的。。。。

class Solution {
public:int maxUncrossedLines(vector<int>& nums1, vector<int>& nums2) {vector<vector<int>> dp(nums1.size()+1, vector<int>(nums2.size()+1, 0));for(int i = 1; i < nums1.size() + 1; i++){for(int j = 1; j < nums2.size() + 1; j++){if(nums1[i-1] == nums2[j-1]){dp[i][j] = dp[i-1][j-1] + 1;}else{dp[i][j] = max(dp[i-1][j], dp[i][j-1]);}}}return dp[nums1.size()][nums2.size()];}
};

第三题、最大子序和 力扣题目链接

class Solution {
public:int maxSubArray(vector<int>& nums) {if(nums.size() == 0) return 0;vector<int> dp(nums.size(), 0);dp[0] = nums[0];int result = nums[0];for(int i = 1; i < nums.size(); i++){dp[i] = max(dp[i-1] + nums[i], nums[i]);if(dp[i] > result) result = dp[i];}return result;}
};