feat:0322重新安排行程 添加JavaScript版本 处理对象key无序问题

This commit is contained in:
eat to 160 pounds
2023-12-08 20:27:07 +08:00
parent 9615a2e2ed
commit 5c20612aba

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@ -562,6 +562,7 @@ func findItinerary(tickets [][]string) []string {
```
### Javascript
```Javascript
var findItinerary = function(tickets) {
@ -605,6 +606,74 @@ var findItinerary = function(tickets) {
```
**javascript版本二 处理对象key无序问题**
```javascript
/**
* @param {string[][]} tickets
* @return {string[]}
*/
var findItinerary = function (tickets) {
const ans = ["JFK"];
let map = {};
// 整理每个站点的终点站信息
tickets.forEach((t) => {
let targets = map[t[0]];
if (!targets) {
targets = { [t[1]]: 0 };
map[t[0]] = targets;
}
targets[t[1]] = (targets[t[1]] || 0) + 1;
});
// 按照key字典序排序对象
const sortObject = (obj) => {
const newObj = {};
const keys = Object.keys(obj);
keys.sort((k1, k2) => (k1 < k2 ? -1 : 1));
keys.forEach((key) => {
if (obj[key] !== null && typeof obj[key] === "object") {
newObj[key] = sortObject(obj[key]);
} else {
newObj[key] = obj[key];
}
});
return newObj;
};
const backtrack = (tickets, targets) => {
if (ans.length === tickets.length + 1) {
return true;
}
const target = targets[ans[ans.length - 1]];
// 没有下一站
if (!target) {
return false;
}
// 或者在这里排序
// const keyList = Object.keys(target).sort((k1, k2) => (k1 < k2 ? -1 : 1));
const keyList = Object.keys(target);
for (const key of keyList) {
// 判断当前站是否还能飞
if (target[key] > 0) {
target[key]--;
ans.push(key);
// 对象key有序 此时的行程就是字典序最小的 直接跳出
if (backtrack(tickets, targets)) {
return true;
}
target[key]++;
ans.pop();
}
}
return false;
};
map = sortObject(map);
backtrack(tickets, map);
return ans;
};
```
### TypeScript
```typescript
@ -901,3 +970,4 @@ impl Solution {
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
</a>