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1.最长公共子序列

1.1 题目

. - 力扣（LeetCode）

1.2 题解

class Solution 
{
public:int longestCommonSubsequence(string text1, string text2) {//确定dp数组，dp[i][j]表示长度为i的text1与长度为j的text2的最长公共子序列的长度vector<vector<int>> dp(text1.size()+1, vector<int>(text2.size()+1, 0));//确定递推逻辑/*if (text1[i - 1] == text2[j - 1])dp[i][j] = dp[i - 1][j - 1] + 1;else dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);*/for (int i = 1; i <= text1.size(); i++){for (int j = 1; j <= text2.size(); j++){if (text1[i - 1] == text2[j - 1])dp[i][j] = dp[i - 1][j - 1] + 1;else dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);}}return dp[text1.size()][text2.size()];}
};

2.不相交的线

2.1 题目

. - 力扣（LeetCode）

2.2 题解

class Solution {
public:int maxUncrossedLines(vector<int>& nums1, vector<int>& nums2) {//确定dp数组，dp[i][j]表示长度为i的nums1与长度为j的nums2的最长公共子序列的长度vector<vector<int>> dp(nums1.size()+1, vector<int>(nums2.size()+1, 0));//确定递推逻辑/*if (nums1[i - 1] == nums2[j - 1])dp[i][j] = dp[i - 1][j - 1] + 1;else dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);*/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][j - 1], dp[i - 1][j]);}}return dp[nums1.size()][nums2.size()];}
};

3.最大子序和

3.1 题目

. - 力扣（LeetCode）

3.2 题解

class Solution {
public:int maxSubArray(vector<int>& nums) {//确定dp数组,dp[i]表示以nums[i]结尾的数组的连续子数组的最大和vector<int> dp(nums.size(), 0);//确定递推关系//dp[i] = max(dp[i - 1] + nums[i],nums[i]);//初始化dp[0] = nums[0];int result = nums[0];//遍历for (int i = 1; i < nums.size(); i++){dp[i] = max(dp[i - 1] + nums[i], nums[i]);result = max(result, dp[i]);}return result;}
};

4.判断子序列

4.1 题目

. - 力扣（LeetCode）

4.2 题解

class Solution {
public:bool isSubsequence(string s, string t) {//确定dp数组,dp[i][j]表示，长度为i的s与长度为j的t的最大公共子序列的长度vector<vector<int>> dp(s.size() + 1, vector<int>(t.size() + 1, 0));//确定递推逻辑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];}}int length= dp[s.size()][t.size()];if (length == s.size())return true;return false;}
};