1143.最长公共子序列
题目链接 文章讲解 视频讲解
- dp[i][j]: 表示以text1以i-1为结尾text2以j-1为结尾的最长公共子序列为dp[i][j]
- 递推公式：如果text1[i-1] == text2[j-1] 那么dp[i][j] = dp[i-1][j-1] + 1;
如果不相同的话，那么dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
class Solution {
public:int longestCommonSubsequence(string text1, string text2) {// dp[i][j]: 表示以text1以i-1为结尾text2以j-1为结尾的最长公共子序列为dp[i][j]vector<vector<int>> dp(text1.size() + 1, vector<int>(text2.size() + 1, 0));// 递推公式 if(text1[i-1] == text2[j-1]) dp[i][j] = dp[i-1][j - 1] + 1;// 初始化// 遍历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]);}}// for(auto vec : dp) {// for(int val : vec) cout << val << " ";// cout << endl;// }return dp[text1.size()][text2.size()];}
};
392.判断子序列
题目链接 文章讲解 视频讲解
class Solution {
public:bool isSubsequence(string s, string t) {if(s.size() > t.size()) return false;vector<vector<int>> dp(s.size() + 1, vector<int>(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] = max(dp[i-1][j], dp[i][j-1]);}}return dp[s.size()][t.size()] == s.size();}
};
