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77 lines
1.6 KiB
Markdown
77 lines
1.6 KiB
Markdown
# [328. Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/)
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## 题目
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Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.
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You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
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**Example 1**:
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```
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Input: 1->2->3->4->5->NULL
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Output: 1->3->5->2->4->NULL
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```
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**Example 2**:
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```
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Input: 2->1->3->5->6->4->7->NULL
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Output: 2->3->6->7->1->5->4->NULL
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```
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**Note**:
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- The relative order inside both the even and odd groups should remain as it was in the input.
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- The first node is considered odd, the second node even and so on ...
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## 题目大意
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这道题和第 86 题非常类型。第 86 题是把排在某个点前面的小值放在一个链表中，排在某个点后端的大值放在另外一个链表中，最后 2 个链表首尾拼接一下就是答案。
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## 解题思路
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这道题思路也是一样的，分别把奇数和偶数都放在 2 个链表中，最后首尾拼接就是答案。
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## 代码
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```go
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package leetcode
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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func oddEvenList(head *ListNode) *ListNode {
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oddHead := &ListNode{Val: 0, Next: nil}
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odd := oddHead
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evenHead := &ListNode{Val: 0, Next: nil}
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even := evenHead
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count := 1
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for head != nil {
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if count%2 == 1 {
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odd.Next = head
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odd = odd.Next
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} else {
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even.Next = head
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even = even.Next
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}
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head = head.Next
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count++
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}
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even.Next = nil
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odd.Next = evenHead.Next
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return oddHead.Next
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}
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``` |