reorder-list

题目描述
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}.

IDEA

1.先找到链表中间点，由此分开两个链表

2.将后边的链表翻转

3.然后将后边链表的每个节点插入第一个链表，进行合并

CODE

/**
 * 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||head->next==NULL)
             return;
        ListNode* mid=findMid(head);
        ListNode* head1=mid->next;
        mid->next=NULL;
        reverseList(head1);   
        merge(head,head1);
    }
    ListNode* findMid(ListNode* head){
        ListNode *fast=head,*slow=head;
        while(fast->next&&fast->next->next){
            slow=slow->next;
            fast=fast->next->next;   
        }
        return slow;
    }
    void reverseList(ListNode* &head){
        if(head==NULL)
            return;
        ListNode* p=head->next;
        ListNode* pre=head;
        ListNode* temp=NULL;
        while(p){
            pre->next = p->next;
            temp=p->next;
            p->next=head;
            head=p;
            p=temp;
        }
    }
    void merge(ListNode* head,ListNode* head1){
        ListNode *p1=head,*p2=head1,*temp = NULL;
        while(p2){
            temp=p2->next;
            p2->next=p1->next;
            p1->next=p2;
            p2=temp;
            p1=p1->next->next;
        }
    }
};