leetcode-143. Reorder List

刚开始刷题，一开始没思路，上网看了一下别人的思路才写出来
总的思路就是先把链表分为两部分，可以先遍历链表再根据长度分也可以用快慢指针（新知识点），然后将第二部分反转，再依次插入到第一部分。思路很简单，但没用ide还是出错了。。。

/**
* 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 ||head->next->next == NULL)
return;
//用快慢指针将链表分为两部分，若为奇数则第一部分大1
ListNode *fast,*slow,*head1,*head2,*post,*cur,*tmp = NULL;
fast = slow = head;
while(fast->next != NULL)
{
fast = fast->next;
if(fast->next == NULL) break;
fast = fast->next;
slow = slow->next;
}
head1 = head;
cur = head2 = slow->next;
slow->next = NULL;  //这里要注意，断开的部分要设NULL
//将第二部分链表反转
post = head2->next;
cur->next = NULL;
while(post!=NULL)
{
tmp = post->next;
post->next = cur;
cur = post;
post = tmp;
}
head2 = cur;
//将第二部分插入到第一部分
ListNode *cur1 = head1;
ListNode *cur2 = head2;
while(cur2!=NULL)
{
tmp = cur1->next;
cur1->next = cur2;
post = cur2->next;
cur2->next = tmp;
cur1 = tmp;
cur2 = post;
}
}
};