143. Reorder List【LeetCode】

问题描述

解题思路：
1.利用快慢指针将链表分成两半
2.将后半段链表逆转（注意，将前半段最后一个指针赋值为null）
3.将两个链表拼接起来

即例如:
1->2->3->4
变成：
1->2
4->3
然后：
1->4->2->3

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) {
        if(!head) return;
        ListNode *slow = head,*fast = head;
	while(fast->next && fast->next->next)
	{
		slow = slow->next;
		fast = fast->next->next;
	}
	
	ListNode *mid = slow->next;
	slow->next = NULL;
	ListNode *pre = NULL; 
	
	while(mid)
	{
		ListNode *temp = mid->next;
		mid->next = pre;
		pre = mid;
		mid = temp;
	}
	while(head && pre)
	{
		ListNode *next = head->next;
		head->next = pre;
		pre = pre->next;
		head->next->next = next;
		head = next;
	}
    }
};