72. 编辑距离
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I am ne zha / Jeskson commented
72. 编辑距离
Description
Difficulty: 困难
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
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]
}