Given a singly linked list L: L0→L1→…→Ln-1→Ln,

reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

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 || !head -> next -> next) return ;        ListNode *slow = head; ListNode *fast = head;        while(fast -> next && fast -> next -> next) {            slow = slow -> next;            fast = fast -> next -> next;        }                ListNode *dummy = slow -> next;        slow -> next = NULL;        dummy = reverseList(dummy);                while(head && dummy) {            ListNode *New = head -> next;            head -> next = dummy;            dummy = dummy -> next;            head -> next -> next = New;            head = New;        }    }        ListNode *reverseList(ListNode *head) {        if(!head || !head -> next) return head;        ListNode *New = reverseList(head -> next);        head -> next -> next = head;        head -> next = NULL;        return New;    }};

　　快慢指针把链表从中间分开然后反转链表 之后把两个链表合并 希望今天出门剪头发之前 Leetcode 刷到 200