mirror of
https://github.com/chefyuan/algorithm-base.git
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178 lines
5.1 KiB
Java
178 lines
5.1 KiB
Java
> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
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>
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号：袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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### [328. 奇偶链表](https://leetcode-cn.com/problems/odd-even-linked-list/)
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下面我们再来看一种双指针，我称之为交替领先双指针，起名鬼才哈哈。下面我们一起来看看吧。
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#### 题目描述
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> 给定一个单链表，把所有的奇数节点和偶数节点分别排在一起。请注意，这里的奇数节点和偶数节点指的是节点编号的奇偶性，而不是节点的值的奇偶性。
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>
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> 请尝试使用原地算法完成。你的算法的空间复杂度应为 O(1)，时间复杂度应为 O(nodes)，nodes 为节点总数。
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示例 1:
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> 输入: 1->2->3->4->5->NULL
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> 输出: 1->3->5->2->4->NULL
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示例 2:
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> 输入: 2->1->3->5->6->4->7->NULL
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> 输出: 2->3->6->7->1->5->4->NULL
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#### 题目解析
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题目也很容易理解就是让我们将原来奇数位的结点放一起，偶数位的结点放一起。这里需要注意，题目和结点值无关，是奇数位和偶数位结点。
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我们可以先将奇数位和在一起，再将偶数位和在一起，最后再将两个链表合并很简单，我们直接看动画模拟吧。
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#### **动画模拟**
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#### 题目代码
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Java Code:
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```java
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class Solution {
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public ListNode oddEvenList(ListNode head) {
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if (head == null || head.next == null) {
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return head;
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}
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ListNode odd = head;
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ListNode even = head.next;
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ListNode evenHead = even;
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while (odd.next != null && even.next != null) {
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//将偶数位合在一起，奇数位合在一起
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odd.next = even.next;
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odd = odd.next;
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even.next = odd.next;
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even = even.next;
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}
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//链接
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odd.next = evenHead;
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return head;
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}
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}
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```
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C++ Code:
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```cpp
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class Solution {
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public:
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ListNode* oddEvenList(ListNode* head) {
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if (head == nullptr || head->next == nullptr) {
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return head;
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}
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ListNode * odd = head;
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ListNode * even = head->next;
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ListNode * evenHead = even;
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while (odd->next != nullptr && even->next != nullptr) {
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//将偶数位合在一起，奇数位合在一起
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odd->next = even->next;
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odd = odd->next;
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even->next = odd->next;
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even = even->next;
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}
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//链接
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odd->next = evenHead;
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return head;
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}
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};
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```
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JS Code:
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```javascript
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var oddEvenList = function (head) {
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if (!head || !head.next) return head;
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let odd = head,
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even = head.next,
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evenHead = even;
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while (odd.next && even.next) {
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//将偶数位合在一起，奇数位合在一起
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odd.next = even.next;
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odd = odd.next;
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even.next = odd.next;
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even = even.next;
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}
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//链接
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odd.next = evenHead;
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return head;
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};
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```
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Python Code:
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```python
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class Solution:
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def oddEvenList(self, head: ListNode) -> ListNode:
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if head is None or head.next is None:
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return head
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odd = head
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even = head.next
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evenHead = even
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while odd.next is not None and even.next is not None:
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# 将偶数位合在一起，奇数位合在一起
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odd.next = even.next
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odd = odd.next
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even.next = odd.next
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even = even.next
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# 链接
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odd.next = evenHead
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return head
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```
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Swift Code：
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```swift
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class Solution {
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func oddEvenList(_ head: ListNode?) -> ListNode? {
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if head == nil || head?.next == nil {
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return head
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}
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var odd = head
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var even = head?.next
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var evenHead = even
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while odd?.next != nil && even?.next != nil {
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//将偶数位合在一起，奇数位合在一起
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odd?.next = even?.next
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odd = odd?.next
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even?.next = odd?.next
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even = even?.next
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}
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//链接
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odd?.next = evenHead
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return head
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}
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}
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```
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Go Code:
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```go
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func oddEvenList(head *ListNode) *ListNode {
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if head == nil || head.Next == nil {
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return head
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}
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odd, even := head, head.Next
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evenHead := even
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for odd.Next != nil && even.Next != nil {
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odd.Next = even.Next
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odd = odd.Next
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even.Next = odd.Next
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even = even.Next
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}
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odd.Next = evenHead
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return head
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}
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```
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