学步园

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

分析

本题思路如下：首先将题目所给单链表从中间分割为两个单链表，再将后半个单链表反向，最后合并两个单链表即可。

代码

public class ReorderList {  
    class ListNode {  
        int val;  
        ListNode next;  
  
        ListNode(int x) {  
            val = x;  
            next = null;  
        }  
    }  
  
    public void reorderList(ListNode head) {  
        if (head == null || head.next == null) {  
            return;  
        }  
  
        // find the second half head  
        ListNode fast = head.next;  
        ListNode slow = head;  
        while (fast != null && fast.next != null) {  
            fast = fast.next.next;  
            slow = slow.next;  
        }  
  
        // reverse the second half  
        ListNode p = slow.next;  
        slow.next = null; // cut the first half  
        ListNode pPre = null;  
        ListNode pSuf = p.next;  
        while (p != null) {  
            pSuf = p.next;  
            p.next = pPre;  
            pPre = p;  
            p = pSuf;  
        }  
  
        // combine two halves  
        ListNode l1 = head;  
        ListNode l2 = pPre;  
        while (l1 != null && l2 != null) {  
            ListNode l1Next = l1.next;  
            ListNode l2Next = l2.next;  
            l1.next = l2;  
            l2.next = l1Next;  
            l1 = l1Next;  
            l2 = l2Next;  
        }  
    }  
}