编辑距离

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

描述

• 给你两个单词 word1 和 word2， 请返回将 word1 转换成 word2 所使用的最少操作数
• 你可以对一个单词进行如下三种操作：
  • 插入一个字符
  • 删除一个字符
  • 替换一个字符

示例 1

输入：word1 = "horse", word2 = "ros"
输出：3

解释：
horse -> rorse (将 ‘h’ 替换为 ‘r’)
rorse -> rose (删除 ‘r’)
rose -> ros (删除 ‘e’)

示例 2

输入：word1 = "intention", word2 = "execution"
输出：5

解释：
intention -> inention (删除 ‘t’)
inention -> enention (将 ‘i’ 替换为 ‘e’)
enention -> exention (将 ‘n’ 替换为 ‘x’)
exention -> exection (将 ‘n’ 替换为 ‘c’)
exection -> execution (插入 ‘u’)

提示

• 0 <= word1.length, word2.length <= 500
• word1 和 word2 由小写英文字母组成

Typescript 版算法实现

1 ) 方案1: 动态规划

function minDistance(word1: string, word2: string): number {const n = word1.length;const m = word2.length;// 有一个字符串为空串if (n * m === 0) {return n + m;}// DP 数组const D: number[][] = Array.from({ length: n + 1 }, () => Array(m + 1).fill(0));// 边界状态初始化for (let i = 0; i < n + 1; i++) {D[i][0] = i;}for (let j = 0; j < m + 1; j++) {D[0][j] = j;}// 计算所有 DP 值for (let i = 1; i < n + 1; i++) {for (let j = 1; j < m + 1; j++) {const left = D[i - 1][j] + 1;const down = D[i][j - 1] + 1;let left_down = D[i - 1][j - 1];if (word1.charAt(i - 1) !== word2.charAt(j - 1)) {left_down += 1;}D[i][j] = Math.min(left, down, left_down);}}return D[n][m];
}

2 ) 方案2: 动态规划自底向上

function minDistance(word1: string, word2: string): number {const n1 = word1.length;const n2 = word2.length;// 初始化 DP 数组const dp: number[][] = Array.from({ length: n1 + 1 }, () => Array(n2 + 1).fill(0));// 初始化第一行for (let j = 1; j <= n2; j++) {dp[0][j] = dp[0][j - 1] + 1;}// 初始化第一列for (let i = 1; i <= n1; i++) {dp[i][0] = dp[i - 1][0] + 1;}// 计算所有 DP 值for (let i = 1; i <= n1; i++) {for (let j = 1; j <= n2; j++) {if (word1.charAt(i - 1) === word2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1];} else {dp[i][j] = Math.min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1;}}}return dp[n1][n2];
}