1143.Longest Common Subsequence
===

###### tags: `Medium`,`String`,`DP`

[1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)

### 題目描述

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.

### 範例

**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.

### 解答

#### Javascript

function longestCommonSubsequence(text1, text2) {
  const m = text1.length;
  const n = text2.length;
  const dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));

  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (text1[i - 1] === text2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1] + 1;
      } else {
        dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
      }
    }
  }

  return dp[m][n];
}

> [name=Marsgoat] [time= Dec 15, 2022]

#### Python

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

        for i in range(1, len(text1)+1):
            for j in range(1, len(text2)+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[m][n]

> [name=Kobe] [time= Dec 15, 2022]
### Reference

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