题意:
给你一个单链表 a1 a2 a3 a4 a5....an 让你变成 a1 an a2 an-1 a3 an-2 ....
这里牵涉到,单链表的倒置和两个单链表的合并。
class Solution {
public:
void reorderList(ListNode *head) {
if (!head || !head->next)return; //空表和只有一个元素的链表不用操作
//快慢指针找中间位置
ListNode *fast = head; ListNode *slow = head;
while (fast&&fast->next)
{
slow = slow->next; fast = fast->next->next;
}
//将后半段的链表倒置
ListNode *after = slow->next; slow->next = NULL;
ListNode *pre = NULL;
while (after){
ListNode *p = after->next;
after->next = pre;
pre = after; after = p;
}
//合并两个链表
ListNode *first = head;
after = pre;
while (first&&after){
ListNode *ftemp = first->next;
ListNode *aftemp = after->next;
first->next = after; first = ftemp;
after->next = first; after = aftemp;
}
}
};