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;
        }
    }
};