思路
快慢指针找到链表的中间结点，将原链表分为两部分，之后将链表的后半部分逆置，之后合并两个单链表，注意：一定要判定头部是否为空！
代码
class Solution {
public:
void reorderList(ListNode *head) {
if (head != NULL){
//快慢指针求中间结点
ListNode *slow = head;
ListNode *fast = head;
while (fast->next != NULL && fast->next->next != NULL){
slow = slow->next;
fast = fast->next->next;
}
//将链表的后面部分逆置
fast = slow->next;
slow->next = NULL;
ListNode *new_head = new ListNode(0);
while (fast){
ListNode *tmp = fast->next;
fast->next = new_head->next;
new_head->next = fast;
fast = tmp;
}
//合并
ListNode *p = head;
ListNode *q = new_head->next;
while (q != NULL){
ListNode *n = q->next;
q->next = p->next;
p->next = q;
p = p->next->next;
q = n;
}
}
}
};
