1143. Longest Common Subsequence

# 1143. Longest Common Subsequence#

## 题目 #

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

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".

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.

## 解题思路 #

• 这一题是经典的最长公共子序列的问题。解题思路是二维动态规划。假设字符串 text1text2 的长度分别为 mn，创建 m+1n+1 列的二维数组 dp，定义 dp[i][j] 表示长度为 i 的 text1[0:i-1] 和长度为 j 的 text2[0:j-1] 的最长公共子序列的长度。先考虑边界条件。当 i = 0 时，text1[] 为空字符串，它与任何字符串的最长公共子序列的长度都是 0，所以 dp[0][j] = 0。同理当 j = 0 时，text2[] 为空字符串，它与任何字符串的最长公共子序列的长度都是 0，所以 dp[i][0] = 0。由于二维数组的大小特意增加了 1，即 m+1n+1，并且默认值是 0，所以不需要再初始化赋值了。

• text1[i−1] = text2[j−1] 时，将这两个相同的字符称为公共字符，考虑 text1[0:i−1]text2[0:j−1] 的最长公共子序列，再增加一个字符（即公共字符）即可得到 text1[0:i]text2[0:j] 的最长公共子序列，所以 dp[i][j]=dp[i−1][j−1]+1。当 text1[i−1] != text2[j−1] 时，最长公共子序列一定在 text[0:i-1], text2[0:j]text[0:i], text2[0:j-1] 中取得。即 dp[i][j] = max(dp[i-1][j], dp[i][j-1])。所以状态转移方程如下：

$dp[i][j] = \left\{\begin{matrix}dp[i-1][j-1]+1 &,text1[i-1]=text2[j-1]\\max(dp[i-1][j],dp[i][j-1])&,text1[i-1]\neq text2[j-1]\end{matrix}\right.$
• 最终结果存储在 dp[len(text1)][len(text2)] 中。时间复杂度 O(mn)，空间复杂度 O(mn)，其中 mn 分别是 text1text2 的长度。

## 代码 #

package leetcode

func longestCommonSubsequence(text1 string, text2 string) int {
if len(text1) == 0 || len(text2) == 0 {
return 0
}
dp := make([][]int, len(text1)+1)
for i := range dp {
dp[i] = make([]int, len(text2)+1)
}
for i := 1; i < len(text1)+1; i++ {
for j := 1; j < len(text2)+1; j++ {
if text1[i-1] == text2[j-1] {
dp[i][j] = dp[i-1][j-1] + 1
} else {
dp[i][j] = max(dp[i][j-1], dp[i-1][j])
}
}
}
return dp[len(text1)][len(text2)]
}

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


Apr 8, 2023