LeetCode 1143 Longest Common Subsequence

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

For example, "ace" is a subsequence of "abcde".
A common subsequence of two strings is a subsequence that is common to both strings.

Solution

求出最长公共子序列。设 \(dp[i][j]\) 表示第一段 \([0,i)\) 和第二段 \([0,j)\) 的答案。考虑如何转移：

如果 \(text1[i-1]=text2[j-1]\):

\[dp[i][j]=1+dp[i-1][j-1] \]

否则，取最大值：

\[dp[i][j]=\max(dp[i][j],\max(dp[i-1][j], dp[i][j-1])) \]

点击查看代码

class Solution {
private:
    int dp[1002][1002];
    // dp[i][j]: [0, i), [0, j)
public:
    int longestCommonSubsequence(string text1, string text2) {
        int n1 = text1.length(), n2 = text2.length();
        for(int i=0;i<=n1;i++){
            for(int j=0;j<=n2;j++)dp[i][j]=INT_MIN;
        }
        for(int i=0;i<=n1;i++){
            for(int j=0;j<=n2;j++){
                if(i==0)dp[i][j]=0;
                if(j==0)dp[i][j]=0;
            }
        }
        for(int i=1;i<=n1;i++){
            for(int j=1;j<=n2;j++){
                if(text1[i-1]==text2[j-1]){
                    dp[i][j] = 1+dp[i-1][j-1];
                }
                else{
                    dp[i][j] = max(dp[i][j], max(dp[i-1][j], dp[i][j-1]));
                }
            }
        }
        return dp[n1][n2];
    }
};