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Recorder List: 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}.
题意：给定一个链表，把最后一个结点插入到第一个结点后面，倒数第二个结点插入到原链表第二个结点后面，依次类推。
思路：利用快慢指针的方法找到链表的中点，然后逆转中点右边的链表，然后将两个链表合并。
代码：
public void reorderList(ListNode head) { if(head == null || head.next == null) return; ListNode slow = head; ListNode fast = head; while(fast.next != null && fast.next.next != null){ slow = slow.next; fast = fast.next.next; } ListNode mid = slow.next; ListNode cur = mid; ListNode pre = null; while(cur != null){ ListNode post = cur.next; cur.next = pre; pre = last; cur = post; } slow.next = null; while(head != null && pre != null){ ListNode next1 = head.next; head.next = pre; pre = pre.next; head.next.next = next1; head = next1; } }
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原文地址：http://www.cnblogs.com/Lewisr/p/5140237.html
