reorder-list——链表、快慢指针、逆转链表、链表合并

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}.

由于链表尾端不干净，导致fast->next!=NULL&&fast->next->next!=NULL判断时仍旧进入循环，此时fast为野指针

1 /** 2 * Definition for singly-linked list.

  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     void reorderList(ListNode *head) {
         if(head==NULL)
             return;
         ListNode *slow=head,*fast=head;//快慢指针找中点
         while(fast->next!=NULL&&fast->next->next!=NULL){
             fast=fast->next->next;
             slow=slow->next;
         }

         ListNode *head1=slow->next;//逆转后半链表
         ListNode *left=NULL;
         ListNode *right=head1;
         while(right!=NULL){
             ListNode *tmp=right->next;
             right->next=left;
             left=right;
             right=tmp;
         }
         head1=left;

         merge(head,head1);//合并两个链表
     }
     void merge(ListNode *left, ListNode *right){
         ListNode *p=left,*q=right;
         while(q!=NULL&&p!=NULL){
             ListNode * nxtleft=p->next;
             ListNode * nxtright=q->next;
             p->next=q;
             q->next=nxtleft;
             p=nxtleft;
             q=nxtright;
         }
     }
 };