leetcode 328 --Odd Even Linked List 链表双指针移动节点

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本文介绍了一种算法，用于将单链表中的奇数节点与偶数节点进行分离，实现奇数节点前置，偶数节点后置的操作。此操作在O(1)空间复杂度和O(nodes)时间复杂度下完成。

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:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->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 …

解题思路：双指针移动，分别存放奇数节点和偶数节点，移动过程中把奇数节点和偶数节点换掉。

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