LeetCode 1143. Longest Common Subsequence

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题目

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:

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

思路

DP.
DP[i][j] represents the longest common subsequence of text1[0 … i] & text2[0 … j].
if text1[i] == text2[j]: DP[i][j] = DP[i - 1][j - 1] + 1
else: DP[i][j] = max(DP[i - 1][j], DP[i][j - 1])

代码

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class Solution(object):
    def longestCommonSubsequence(self, text1, text2):
        """
        :type text1: str
        :type text2: str
        :rtype: int
        """
        l1, l2 = len(text1), len(text2)
        dp = [[0 for _ in range(l2+1)] for _ in range(l1+1)]
        for i in range(1, l1+1):
            for j in range(1, l2+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[l1][l2]