Given a singly linked listL: 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}.

解题思路:

    用栈将整个链表存储，然后将栈顶元素插入到栈底元素的后面，循环多少次呢？链表

的长度除2即可.

解题代码:

/** * 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)    {        stack<ListNode *> stk;        ListNode *tmp = head;        int cnt = 0 ;        while(tmp)        {            stk.push(tmp);            tmp = tmp->next;            ++cnt;        }        tmp = head ;        for(int i = 1 ; i <= cnt / 2 ; ++i)        {            ListNode *tmp1 = stk.top();            stk.pop();            tmp1->next = tmp->next ;            tmp->next = tmp1 ;            tmp = tmp1->next;        }        if(head)            tmp->next = NULL ;    }};

  注：上述代码的空间复杂度是O(n),不过这题应该可以做到O(1)的空间的,遍历链表从中间分割即可.