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→L2→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: 这个就是先用快慢指针把链表进行拆开,分成两个,然后把后面那部分链表进行反转,最后再把两条链表进行拼接
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)//step1
return ;
ListNode slow=head;
ListNode fast=head;
while(fast.next!=null&&fast.next.next!=null){
slow=slow.next;
fast=fast.next.next;
}
ListNode nextHead=slow.next;//千万注意 进行到这里的时候 必须至少有两个节点,所以step1处必须先对head.next进行判空
slow.next=null;
ListNode first=new ListNode(0);
first.next=nextHead;
ListNode cur2=nextHead;
while(cur2.next!=null){
ListNode temp=cur2.next;
cur2.next=temp.next;
temp.next=first.next;
first.next=temp;
}
ListNode cur1=head;
cur2=first.next;
while(cur2!=null){
ListNode temp1=cur1.next;
ListNode temp2=cur2.next;
cur1.next=cur2;
cur2.next=temp1;
cur1=temp1;
cur2=temp2;
}
}
}