reorder-list——链表快慢指针逆转链表链表合并

Posted 2020-09-21 鸭子船长

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.

 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void reorderList(ListNode *head) {
12         if(head==NULL)
13             return;

14         ListNode *slow=head,*fast=head;//快慢指针找中点
15         while(fast->next!=NULL&&fast->next->next!=NULL){
16             fast=fast->next->next;
17             slow=slow->next;
18         }
19         
20         ListNode *head1=slow->next;//逆转后半链表
21         ListNode *left=NULL;
22         ListNode *right=head1;
23         while(right!=NULL){
24             ListNode *tmp=right->next;
25             right->next=left;
26             left=right;
27             right=tmp;
28         }
29         head1=left;
30         
31         merge(head,head1);//合并两个链表
32     }
33     void merge(ListNode *left, ListNode *right){
34         ListNode *p=left,*q=right;
35         while(q!=NULL&&p!=NULL){
36             ListNode * nxtleft=p->next;
37             ListNode * nxtright=q->next;
38             p->next=q;
39             q->next=nxtleft;
40             p=nxtleft;
41             q=nxtright;
42         }
43     }
44 };