思路

dp[i][j] 表示 text1[i-1] 和text2[j-1] 的最长公共子序列，dp[0][0] = 0 ,初始化为0；如果text1[i-1] == text2[j-1],那么dp[i][j] = dp[i-1][j-1]+1 ; 如果不相等，那么dp[i][j] = max(dp[i-1][j],dp[i][j-1]);

仔细体会

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int len1 = text1.length();
        int len2 = text2.length();
        vector<vector<int>> dp(len1+1,vector<int>(len2+1,0));
        for(int i = 1 ; i< len1+1;i++){
            for(int j = 1;j<len2+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[len1][len2];
    }
};