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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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增加 0583 两个字符串的删除操作 Javascript 方法二
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wang2jun
2022-11-08 15:01:42 +08:00
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commit d5392f4936
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problems/0583.两个字符串的删除操作.md
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@ -228,28 +228,43 @@ func min(a, b int) int {
```
Javascript：
```javascript
const minDistance = (word1, word2) => {
let dp = Array.from(new Array(word1.length + 1), () => Array(word2.length+1).fill(0));
for(let i = 1; i <= word1.length; i++) {
dp[i][0] = i;
// 方法一
var minDistance = (word1, word2) => {
let dp = Array.from(new Array(word1.length + 1), () =>
Array(word2.length + 1).fill(0)
);
for (let i = 1; i <= word1.length; i++) {
dp[i][0] = i;
}
for (let j = 1; j <= word2.length; j++) {
dp[0][j] = j;
}
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];
} else {
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];
};
for(let j = 1; j <= word2.length; j++) {
dp[0][j] = j;
}
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];
} else {
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];
// 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;
};
```