[LeetCode] Reorder List

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n, reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-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}.

/**
 * 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) {
        ListNode *p1=head,*p2=head,*pre;
        while(p2!=NULL && p2->next!=NULL)//找到后一半的开头节点head2；
        {
          pre = p1;
          p1 = p1->next;
          p2 = p2->next;
          p2 = p2->next;
        }
        
        if(p1 == p2)
            return;
        pre->next = NULL;//前一半的结尾标记
        ListNode *head2 = p1;
        
       stack<ListNode*> st;
       while(head2)
       {
         st.push(head2);
         head2 = head2->next;
       }
       p1 = head;
       ListNode *ph1,*ph2;
       while(p1 && !st.empty() )
       {
         ph1 = p1->next;
         ph2 = st.top();
         st.pop();
         p1->next = ph2;
         ph2->next = ph1;
         p1 = ph1;
       
       }
       if(!st.empty())
       {
         ph1 = st.top();
         ph1->next = NULL;
         st.pop();
         ph2->next = ph1;
       
       }

    }
};

（1）注意数据结构用stack比较方便，如果用遍历的方法，则会超时Time Limited Exceeded！
 （2）测试用例中，head包括NULL,1个结点、2个结点、3个结点、4个结点，把这几个测试正确，其余的应该就没有问题了。