leetcode-1143-Longest-Common-Subsequence

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描述

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:

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

Example 2:

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

Example 3:

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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.

分析

dp[i][j] 表示 text1 的前 i 个字符和 text2 的前 j 个字符，当 text1[i-1]==text2[j-1], dp[i][j]=dp[i-1][j-1]+1, 否则 dp[i][j] = max(dp[i-1][j], dp[i][j-1])

解决方案1(Python)

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class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        text1_length, text2_length = len(text1), len(text2)
        
        dp = [[0] * (text2_length+1) for _ in range(text1_length+1)]
        
        for i in range(1, text1_length+1):
            for j in range(1, text2_length+1):
                if 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[text1_length][text2_length]

解决方案2（Golang）

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func longestCommonSubsequence(text1 string, text2 string) int {
    text1_length, text2_length := len(text1), len(text2)
    dp := make([][]int, text1_length+1)
    for i := range dp {
        dp[i] = make([]int, text2_length+1)
    }
    
    for i, ch1 := range(text1) {
        for j, ch2 := range(text2) {
            if ch1 == ch2 {
                dp[i+1][j+1] = dp[i][j] + 1
            } else {
                dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
            }
        }
    }
    return dp[text1_length][text2_length]
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

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