1035.不相交的线 

其实就是和1143一模一样，只要顺序相同，就可以实现不相交。

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

一维数组的思路很难其实，暂且放一下，先做下一题。

#include <iostream>
#include <string>
#include <vector>
using namespace ::std;
class Solution
{
public:int longestCommonSubsequence(string text1, string text2){// 首先要用text2的长度vector<int> dp(text2.size() + 1, 0);for (int i = 1; i <= text1.size(); i++){int pre = dp[0];for (int j = 1; j <= text2.size(); j++){int cur = dp[j];if (text1[i - 1] == text2[j - 1]){dp[j] = pre + 1;}else{dp[j] = max(dp[j], dp[j - 1]);}pre = cur;}cout << i << endl;for (int j = 0; j <= text2.size(); j++){cout << dp[j] << ' ';}cout << endl;}return dp[text2.size()];}
};
int main()
{Solution syz;string text1 = "abcde";string text2 = "ace";syz.longestCommonSubsequence(text1, text2);return 0;
}

1
0 1 1 1
2
0 1 1 1
3
0 1 2 2
4
0 1 2 2
5
0 1 2 3

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