**Leetcode_reorder-list

地址：http://oj.leetcode.com/problems/reorder-list/

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

reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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}

.
思路：感觉这是一道非常好的题，有比较多基础的知识点。比如两个指针一个步进为一，另一个步进为二来求链表的中间节点，还有就是reverseList是in-place的反转单链表，最后就是把两个长度差小于1的链表锯齿形的merge在一起。
把每个知识点都好好想了一想，这样的题应该多做，打上两颗*.

参考代码：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head)
{
ListNode*pre=NULL, *cur=head, *next=NULL;
while(cur)
{
next=cur->next;
cur->next=pre;
pre=cur;
cur=next;
}
return pre;
}
void reorderList(ListNode *head) {
if(head)
{
ListNode *p1 = head, *p2 = head, *nxt1 = NULL, *nxt2 = NULL;
while(p2 && p2->next)
{
p1 = p1->next;
p2 = p2->next;
if(p2)
p2 = p2->next;
}
p2 = p1->next;
p1->next = NULL;
p1 = head;
p2 = reverseList(p2);
//merge
while(p1 && p2)
{
nxt1 = p1->next;
nxt2 = p2->next;
p1->next = p2;
p2->next = nxt1;
p1 = nxt1;
p2 = nxt2;
}
}
}
};

//SECOND TRIALclass Solution {private:    ListNode* reverseList(ListNode* head)    {        if(!head || !head->next)            return head;        ListNode *p = head, *pre = NULL, *nxt = NULL;        while(p)        {            nxt = p->next;            p->next = pre;            pre = p;            p = nxt;        }        return pre;    }public:    void reorderList(ListNode *head) {        if(!head || !head->next)            return;        ListNode*p1 = head, *p2 = head, *pre = NULL, *nxt1 = NULL, *nxt2 = NULL;        while(p1)        {            p1 = p1->next;            pre = p2;            if(p1)            {                p1 = p1->next;                p2 = p2->next;            }        }        p2 = pre->next;        pre->next = NULL;        p2 = reverseList(p2);        p1 = head;        while(p1 && p2)        {            nxt1 = p1->next;            nxt2 = p2->next;            p1->next = p2;            p1 = p2->next = nxt1;            p2 = nxt2;        }    }};