【LeetCode】#143重排链表(Reorder List)

【LeetCode】#143重排链表(Reorder List)

题目描述

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…
你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例

示例 1:

给定链表 1->2->3->4, 重新排列为 1->4->2->3.

示例 2:

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

Description

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You may not modify the values in the list’s nodes, only nodes itself may be changed.

Example

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

解法

class Solution {
    public void reorderList(ListNode head) {
        if(head==null || head.next==null || head.next.next==null){
            return ;
        }
        ListNode slow = head;
        ListNode fast = head;
        while(fast!=null && fast.next!=null){
            slow = slow.next;
            fast = fast.next.next;
        }
        fast = slow.next;
        slow.next = null;
        ListNode pre = null, next = null;
        while(fast!=null){
            next = fast.next;
            fast.next = pre;
            pre = fast;
            fast = next;
        }
        fast = pre;
        while(head!=null && fast!=null){
            ListNode tmp = head.next;
            ListNode tmp2 = fast.next;
            head.next = fast;
            fast.next = tmp;
            head = tmp;
            fast = tmp2;
        }
    }
}