Leetcode- 143. Reorder List

解题思路：使用快慢指针找到链表的中间位置（代码需要写的非常小心）。对于后半部分的链表逆序（如果不使用哑节点dummy方法，需要在之前判断链表元素个数小于1时直接返回。）

/**
 * 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==NULL || head->next==NULL)
            return;
        ListNode *slow=head,*fast=head->next;
        while(fast && fast->next){
            fast = fast->next->next;
            slow = slow->next;
        }

        
        ListNode *newh = slow->next;
        slow->next = NULL;        
        fast = newh->next;
        newh->next =NULL;
        
        while(fast){
            ListNode *tmp  = fast->next;
            fast->next = newh;
            newh = fast;
            fast = tmp;
        }
        
        ListNode *cur = head;
        while(cur && newh){
            ListNode *post = cur->next;
            ListNode *tmp = newh;
            newh = newh->next;
            cur->next = tmp;
            tmp->next = post;
            cur = post;
        }
    }
};