leetcode之Reorder List

原题如下：

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}.

这道题难度不算太大，实现主要分为如下几步：首先是利用快慢指针找到中间节点，然后从中间节点将链表断开为两个链表（总节点为奇数个时第一个链表多一个节点），然后将第二个链表（后半部分链表）进行逆转，然后将第一个链表和逆转后的第二个链表进行合并，这里要注意对第一个链表多一个节点的情况的处理。

class Solution {
public:
    void reorderList(ListNode *head) {
		if(head == NULL || head->next == NULL)
			return;
		ListNode *fast = head, *slow = head;
		while(fast->next != NULL && fast->next->next != NULL){
			fast = fast->next->next;
			slow = slow->next;
		}
		fast = slow->next;
		slow->next = NULL;
		fast = reverse(fast);
	    merge(head,fast);
		return;
    }
	ListNode* reverse(ListNode *head){
		if(head == NULL || head->next == NULL)
			return head;
		ListNode *p = head->next,*post = head->next;
		head->next = NULL;
		while(p != NULL){
			post = p->next;
			p->next = head;
			head = p;
			p = post;
		}
		return head;
	}
	ListNode *merge(ListNode *head1, ListNode *head2){
		ListNode *head = new ListNode(0);
		ListNode *p = head;
		while(head1 != NULL && head2 != NULL){
			p->next = head1;
			head1 = head1->next;
			p->next->next = head2;
			p = head2;
			head2 = head2->next;			
		}
		if(head1 != NULL)
			p->next = head1;
		else 
			p->next = head2;
		head1 = head->next;
		delete head;
		return head1;
	}
};