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Merge pull request #1743 from wang2jun/master

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增加 0583 两个字符串的删除操作 Javascript 方法二
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程序员Carl
2022-11-21 11:52:54 +08:00
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parent ef4cf1a07b a6f2d48d8a
commit 191d6f3eed
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problems/0583.两个字符串的删除操作.md
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@ -228,29 +228,44 @@ func min(a, b int) int {
``` ```
Javascript： Javascript：
```javascript ```javascript
const minDistance = (word1, word2) => { // 方法一
let dp = Array.from(new Array(word1.length + 1), () => Array(word2.length+1).fill(0)); var minDistance = (word1, word2) => {
let dp = Array.from(new Array(word1.length + 1), () =>
for(let i = 1; i <= word1.length; i++) { Array(word2.length + 1).fill(0)
);
for (let i = 1; i <= word1.length; i++) {
dp[i][0] = i; dp[i][0] = i;
} }
for (let j = 1; j <= word2.length; j++) {
for(let j = 1; j <= word2.length; j++) {
dp[0][j] = j; dp[0][j] = j;
} }
for (let i = 1; i <= word1.length; i++) {
for(let i = 1; i <= word1.length; i++) { for (let j = 1; j <= word2.length; j++) {
for(let j = 1; j <= word2.length; j++) { if (word1[i - 1] === word2[j - 1]) {
if(word1[i-1] === word2[j-1]) { dp[i][j] = dp[i - 1][j - 1];
dp[i][j] = dp[i-1][j-1];
} else { } else {
dp[i][j] = Math.min(dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + 2); dp[i][j] = Math.min(
dp[i - 1][j] + 1,
dp[i][j - 1] + 1,
dp[i - 1][j - 1] + 2
);
} }
} }
} }
return dp[word1.length][word2.length]; return dp[word1.length][word2.length];
}; };
// 方法二
var minDistance = function (word1, word2) {
let dp = new Array(word1.length + 1)
.fill(0)
.map((_) => new Array(word2.length + 1).fill(0));
for (let i = 1; i <= word1.length; i++)
for (let j = 1; j <= word2.length; j++)
if (word1[i - 1] === word2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
return word1.length + word2.length - dp[word1.length][word2.length] * 2;
};
``` ```
TypeScript： TypeScript：