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}.

借助栈可以实现时间复杂度为O(N),找到链表中点，那么从中点往左的每一个结点都是需要插入结点的点，而从中点往右的结点都是需要被插入到前半部分的结点。代码如下：

/**
 * 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) {
        stack<ListNode *> paths;
        int len = length(head);
        if(len<=2) return;
        int repeatTimes = len%2 ? len/2:len/2-1;
        int i=1;
        ListNode *p=head;
        while(i <= repeatTimes){
            paths.push(p);
            if(i!=repeatTimes)
               p=p->next;
            ++i;
        }
        ListNode *pre = p;//此处的p是栈顶元素
        if(len%2){//奇数的情况
            pre = p->next;
        }else{
            pre = p->next->next;
        }
        p = pre->next;
        while(p){
            pre->next = p->next ;
            if(!paths.empty()){
                ListNode *q = paths.top();
                p->next = q->next;
                q->next = p;
                paths.pop();
                p = pre->next;
            }
        }
    }
private:
   int length(ListNode *head){
       int len = 0 ;
       while(head){
           ++len;
           head=head->next;
       }
       return len;
   }
};