leetcode 1143. 最长公共子序列-优快云博客

本文链接：https://blog.youkuaiyun.com/m0_37454852/article/details/115413888

本文解析了如何使用动态规划解决最长公共子序列问题，通过实例展示了Solution类中两种不同初始化方式的vector dp数组，详细讲解了状态转移方程和求解过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

在这里插入图片描述

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

在这里插入图片描述

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