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feat: update js solution to lcof problem: No.25 #708

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merged 1 commit into from
Feb 11, 2022
Merged

feat: update js solution to lcof problem: No.25 #708

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66 changes: 40 additions & 26 deletions lcof/面试题25. 合并两个排序的链表/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -79,6 +79,8 @@ class Solution {

### **JavaScript**

- 递归

```js
/**
* Definition for singly-linked list.
Expand All @@ -93,39 +95,51 @@ class Solution {
* @return {ListNode}
*/
var mergeTwoLists = function (l1, l2) {
// 法一 - 递归
if (!l1) return l2;
if (!l2) return l1;
if (!(l1 && l2)) {
return l1 || l2;
}
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l2.next, l1);
return l2;
}
// 法二 - 遍历
// if(!l1 || !l2) return l1 ? l1 : l2
// let a = l1
// let b = l2
// let res = l1
// if(a.val > b.val) {
// let c = a
// a = b
// b = c
// res = l2
// }
// while(a && b) {
// while(a.next && a.next.val < b.val) {
// a = a.next
// }
// let tmp = a.next
// let rec = b.next
// a.next = b
// a.next.next = tmp
// a = a.next
// b = rec
// }
// return res
};
```

- 遍历

```js
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var mergeTwoLists = function(l1, l2) {
const res = new ListNode();
let cur = res;
while (l1 && l2) {
let node;
if (l1.val < l2.val) {
node = l1;
l1 = l1.next;
} else {
node = l2;
l2 = l2.next;
}
cur.next = node;
cur = node;
}
cur.next = l1 || l2;
return res.next;
};
```

Expand Down
29 changes: 3 additions & 26 deletions lcof/面试题25. 合并两个排序的链表/Solution.js
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Original file line number Diff line number Diff line change
Expand Up @@ -11,37 +11,14 @@
* @return {ListNode}
*/
var mergeTwoLists = function (l1, l2) {
// 法一 - 递归
if (!l1) return l2;
if (!l2) return l1;
if (!(l1 && l2)) {
return l1 || l2;
}
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l2.next, l1);
return l2;
}
// 法二 - 遍历
// if(!l1 || !l2) return l1 ? l1 : l2
// let a = l1
// let b = l2
// let res = l1
// if(a.val > b.val) {
// let c = a
// a = b
// b = c
// res = l2
// }
// while(a && b) {
// while(a.next && a.next.val < b.val) {
// a = a.next
// }
// let tmp = a.next
// let rec = b.next
// a.next = b
// a.next.next = tmp
// a = a.next
// b = rec
// }
// return res
};