72. 编辑距离 @Python @Go

原题

https://leetcode.cn/problems/edit-distance/description/

思路

动态规划

复杂度

时间：O(m * n)
空间：O(m * n)

Python代码

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        # dp[i][j] 表示从 word1 的前i个字符转换到 word2 的前j个字符所需要的步骤
        dp = [[0] * (n + 1) for i in range(m + 1)]
        for i in range(m + 1):
            dp[i][0] = i 
        for j in range(n + 1):
            dp[0][j] = j 
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1 
        return dp[-1][-1]
        

Go代码

func minDistance(word1 string, word2 string) int {
	m, n := len(word1), len(word2)
	// dp[i][j] 表示从 word1 的前i个字符转换到 word2 的前j个字符所需要的步骤
	dp := make([][]int, m+1)
	for i := 0; i < m+1; i++ {
		dp[i] = make([]int, n+1)
	}
	for i := 0; i < m+1; i++ {
		dp[i][0] = i
	}
	for j := 0; j < n+1; j++ {
		dp[0][j] = j
	}
	for i := 1; i < m+1; i++ {
		for j := 1; j < n+1; j++ {
			if word1[i-1] == word2[j-1] {
				dp[i][j] = dp[i-1][j-1]
			} else {
				dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1
			}
		}
	}
	return dp[m][n]
}