Description

Given two strings text1 and text2, return the length of their longest common subsequence.

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. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A common subsequence of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

Example 1:

Input: text1 = "abcde", text2 = "ace" 
Output: 3  
Explanation: The longest common subsequence is "ace" and its length is 3.

Example 2:

Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.

Example 3:

Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

Constraints:

  • 1 <= text1.length <= 1000
  • 1 <= text2.length <= 1000
  • The input strings consist of lowercase English characters only.

分析

题目的意思是：求两个字符串的最长公共子序列，不要求连续。这道题比较中规中矩，做了好多遍了，dp[i][j]表示text 0…i-1, text2 0…j-1的最长公共子序列，递推公式也比较简单，分两种情况，当前的字符串相等：

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

当前的字符串不相等：

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

代码

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m=len(text1)
        n=len(text2)
        dp=[[0]*(n+1) for i in range(m+1)]
        for i in range(m+1):
            for j in range(n+1):
                if(i==0 or j==0):
                    continue
                elif(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[m][n]