leetcode题目4：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}

.

题意很好理解，就是讲链表按照顺序重新排序，我的思路是利用两个指针，一个指向表头，一个指向表尾，表头先入栈然后指向下一个元素，然后表尾入栈，指向倒数第二个元素，做循环直至两个指针相遇，此时栈中的元素应该是排好序的

AC代码：

/**
* 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||!head->next)return;
list<ListNode*>order,reorder;
ListNode *p=head;
for(p;p;p=p->next)
order.push_back(p);
list<ListNode*>::iterator front,back;
front=order.begin();
back=--order.end();
while(front!=back)
{
if(front!=back)
{
reorder.push_back(*front);
front++;
}
if(front!=back)
{
reorder.push_back(*back);
back--;
}
}
reorder.push_back(*front);
front=reorder.begin();
ListNode *q=*front;
p=q;
for(front++;front!=reorder.end();front++)
{
p->next=*front;
p=p->next;
}
p->next=NULL;
head=q;
}
};