LeeCode 328 Odd Even Linked List (链表)

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本文介绍了一种在O(1)空间复杂度和O(nodes)时间复杂度下，将单链表中的奇数节点和偶数节点分别组合在一起的算法。通过一个简单的迭代过程，该算法保持了原始输入中奇数和偶数节点的相对顺序。

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.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL

Example 2:

Input:2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL

Note:

The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on ...

题目链接：https://leetcode.com/problems/odd-even-linked-list/

题目分析：奇偶交错模拟即可

0ms，时间击败100%，空间击败100%

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    
    public ListNode oddEvenList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode odd = head;
        ListNode even = head.next;
        ListNode evenHead = even;
        while (even != null && even.next != null) {
            odd.next = even.next;
            odd = odd.next;
            even.next = odd.next;
            even = even.next;
        }
        odd.next = evenHead;
        return head;
    }
}