题解 | #计算字符串的编辑距离#

package main

import "fmt"

func solution(str1, str2 string) int {
	dp := make([][]int, len(str1)+1)
	for i := range dp {
		dp[i] = make([]int, len(str2)+1)
	}
	//初始化dp
	for i := 1; i < len(dp); i++ {
		dp[i][0] = dp[i-1][0] + 1
	}
	for i := 1; i < len(dp[0]); i++ {
		dp[0][i] = dp[0][i-1] + 1
	}
	//开始dp
	for i := 0; i < len(str1); i++ {
		for j := 0; j < len(str2); j++ {
			if str1[i] == str2[j] {
				dp[i+1][j+1] = dp[i][j]
			} else {
				dp[i+1][j+1] = min(dp[i][j], min(dp[i+1][j], dp[i][j+1])) + 1
			}
		}
	}
	return dp[len(str1)][len(str2)]
}

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

/*
*
//结果3
str1 := "abcdefg"
str2 := "abcd"

//结果3
str1 := "horse"
str2 := "ros"
*/
func main() {
	var str1, str2 string
	fmt.Scanln(&str1)
	fmt.Scanln(&str2)
	fmt.Println(solution(str1, str2))
}