1143.Longest_Common_Subsequence.py

"""
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.

 

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, text2.length <= 1000
text1 and text2 consist of only lowercase English characters.
"""

"""
1. dp[i][j]为前i个字符A[0,i)和前j个字符[0...j)的最长公共字串的长度，注意不包括i,j
2. dp[m+1][n+1]
3. dp
4. dp[i][j] ifA[i]=B[j] dp[i][j] = dp[i-1][j-1] + 1 else dp[i][j] = max(dp[i+1][j], dp[i][j-1])
"""
class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m = len(text1)
        n = len(text2)
        
        dp = [[0]*(n+1) for _ in range(m+1)]
        
        for i in range(1, m+1):
            for j in range(1, n+1):
                if text1[i-1] == text2[j-1]:    #注意这里容易出错，dp[i][j]是前i个，所以应该比较text1[i-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]