题目描述:
Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-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}.
思路解析:
由题意可知:
- 首先把链表分为两半,firstlist和secondlist,然后把secondlist逆序,然后穿插就可以了
- 注意分两半的时候把slow.next=null
- 还要注意反转的时候要提前把cur的next先保存一下
- 还要注意穿插时,只需要截至到temp1就可以,然后temp1又代表firstlist,而temp2又代表secondlist
- 注意逆序时,要把pre(前面的)插入到cur(原来是后边的)的后边
代码:
/** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public void reorderList(ListNode head) { if(head ==null||head.next==null) return; ListNode slow = head,fast = head; ListNode firstlist = head; while(fast.next!=null&&fast.next.next!=null){ slow = slow.next; fast = fast.next.next; } ListNode secondlist =slow.next; slow.next = null; secondlist = reverseOrder(secondlist); while(secondlist!=null){ ListNode temp1 = firstlist.next; ListNode temp2 = secondlist.next; firstlist.next = secondlist; secondlist.next = temp1; firstlist =temp1; secondlist = temp2; } } public ListNode reverseOrder(ListNode head){ if(head == null) return head; ListNode pre = head; ListNode cur = head.next; while(cur!=null){ ListNode temp = cur.next; cur.next = pre; pre = cur; cur = temp; } head.next= null; return pre; } }