LeetCode Hot 100 P1143 Longest Common Subsequence

---

Timeline

Timeline

2026-03-21

init

Dynamic Programming

---

Problem:

P1143 Longest Common Subsequence

https://leetcode.cn/problems/longest-common-subsequence/description/?envType=study-plan-v2&envId=top-100-liked

This is a classic problem,dp[i][j]representstext1[0..=i]andtext2[0..=j]the length of the longest common subsequence, and the state transition equation is

• When $text1[i] == text2[j]$:
  • $dp[i][j] = dp[i-1][j-1] + 1$
• otherwise:
  • $dp[i][j] = max(dp[i-1][j], dp[i][j-1])$

#include <string>
#include <vector>

using std::string;
using std::vector;

class Solution {
    public:
        int longestCommonSubsequence(string text1, string text2)
        {
                int i, j, n1 = text1.size(), n2 = text2.size();
                vector<vector<int> > dp(n1, vector<int>(n2, 0));

                for (i = 0; i < n1; i++) {
                        if (text1[i] == text2[0]) {
                                while (i < n1)
                                        dp[i++][0] = 1;
                                break;
                        }
                }

                for (j = 0; j < n2; j++) {
                        if (text2[j] == text1[0]) {
                                while (j < n2)
                                        dp[0][j++] = 1;
                                break;
                        }
                }

                // dp[i][j] represents the length of the longest common subsequence of text1[0..=i] and text2[0..=j]

                // dp[i][j] = dp[i-1][j-1] + 1 (text1[i] == text2[j])
                // dp[i][j] = max(dp[i-1][j], dp[i][j-1])
                for (i = 1; i < n1; i++) {
                        for (j = 1; j < n2; j++) {
                                if (text1[i] == text2[j])
                                        dp[i][j] = dp[i - 1][j - 1] + 1;
                                else
                                        dp[i][j] = std::max(dp[i - 1][j], dp[i][j - 1]);
                        }
                }

                return dp[n1 - 1][n2 - 1];
        }
};