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}.

思路：用队列或者栈将整个链表存储起来，并计算链表长度。然后将队尾或者栈顶元素取出插入到对头元素的后面。这样就可以完成要求了。

/**
 * 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) {
        deque<ListNode *> pList;
        ListNode *pNode=head;
        ListNode *pCur=head;
        int count=0;
        while(pNode!=NULL)
        {
            pList.push_back(pNode);
            pNode=pNode->next;
            count++;
        }
        for(int i=0;i<count/2;i++)
        {
            ListNode *temp=pList.back();
            pList.pop_back();
            temp->next=pCur->next;
            pCur->next=temp;
            pCur=temp->next;
        }
        if(head!=NULL)
            pCur->next=NULL;
    }
};