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72. 编辑距离 #30

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webVueBlog opened this issue Sep 4, 2022 · 0 comments
Open

72. 编辑距离 #30

webVueBlog opened this issue Sep 4, 2022 · 0 comments

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@webVueBlog
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72. 编辑距离

Description

Difficulty: 困难

Related Topics: 字符串, 动态规划

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
  • word1word2 由小写英文字母组成

Solution

Language: JavaScript

/**
 * @param {string} word1
 * @param {string} word2
 * @return {number}
 */
// dp[i][j]是word1,word2的最小编辑距离
var minDistance = function(word1, word2) {
    let m = word1.length
    let n = word2.length
    const dp = new Array(m+1).fill(0).map(() => new Array(n+1).fill(0))
    // 初始化
    for (let i = 1; i <= m; i++) {
        dp[i][0] = i
    }
    for (let j = 1; j <= n; j++) {
        dp[0][j] = j
    }
    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            if (word1[i-1] === word2[j-1]) {
                dp[i][j] = dp[i-1][j-1]
            } else {
                dp[i][j] = Math.min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
            }
        }
    }
    return dp[m][n]
}
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