leetcode 143. Reorder List

143. Reorder List

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}.

（法1）我的思路是每次找最后一个，然后插入到应该的位置，但是这样为o(n^2)。

（法2）还有一个思路是中间分为两节，后一节倒置，然后挨个插入,o(n)。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* Reverse2(ListNode* head)
    {
        if(!head)
            return head;
        else if(!head->next)
            return head;
        else
        {
            ListNode* tail=head; 
            ListNode* coming=head->next;
          
            while(coming)
            {
                tail->next=coming->next;
                coming->next=head;
              
                head=coming;
                coming=tail->next;
            }
            return head; 
        }
        return head;
    }
    
    void reorderList(ListNode* head) 
    {
        if(!head)
            return;   
        
        ListNode *fast=head;
        ListNode *slow=head;
       
        //将链表拆为两节
        while(fast && fast->next)
        {
            fast=fast->next->next;
            slow=slow->next;
        }
        fast=slow->next;
        slow->next=NULL;
       
        //转置第二节
        fast=Reverse2(fast);
       
        //合并
        ListNode *p=fast;
        ListNode *q=head;
        while(fast && q)
        {
            p=fast;
            fast=fast->next;
          
            p->next=q->next; 
            q->next=p;
            q=p->next;      
        }
    }
};