72. 编辑距离

最新推荐文章于 2025-02-02 02:21:21 发布

原创最新推荐文章于 2025-02-02 02:21:21 发布 · 91 阅读

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题目.

难度:

high

解答：

乍一看无从下手，可是我们有万能的动态规划。推导公式就是，当目标字符串增加一个字符的时候需要怎么根据之前的结果得到当前值。完全就是动态规划经典问题：寻找最长公共字符串的翻版

package main

import "fmt"

func minDistance(word1 string, word2 string) int {
	if len(word1) == 0 {
		return len(word2)
	}

	if len(word2) == 0 {
		return len(word1)
	}

	dp := make([][]int, len(word1)+1)
	for i := 0; i <= len(word1); i++ {
		dp[i] = make([]int, len(word2)+1)
	}

	for i := 1; i <= len(word1); i++ {
		dp[i][0] = i
	}

	for j := 1; j <= len(word2); j++ {
		dp[0][j] = j
	}

	for i := 1; i <= len(word1); i++ {
		for j := 1; j <= len(word2); j++ {
			//add
			dp[i][j] = dp[i][j-1] + 1

			//match
			if word1[i-1] == word2[j-1] && dp[i-1][j-1] < dp[i][j] {
				dp[i][j] = dp[i-1][j-1]
			}

			//replace
			if dp[i-1][j-1]+1 < dp[i][j] {
				dp[i][j] = dp[i-1][j-1] + 1
			}

			//discard
			if dp[i-1][j]+1 < dp[i][j] {
				dp[i][j] = dp[i-1][j] + 1
			}

		}
	}

	//fmt.Println(dp)
	return dp[len(word1)][len(word2)]
}

func main() {
	fmt.Println(minDistance("horse", "ros"))
}