题意:
原题链接
思路:
- 快慢指针找到中点(或者先遍历得到长度,再遍历一半也可行)
- 反转后半部分
- 归并两部分
代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head == NULL || head->next == NULL) return head;
ListNode *next_tail = head->next;
ListNode *next_head = reverseList(head->next);
next_tail->next = head;
head->next = NULL;
return next_head;
}
void reorderList(ListNode* head) {
ListNode *dummy = new ListNode(-1);
dummy->next = head;
ListNode *quick = dummy, *slow = dummy;
while(quick) {
if(quick) quick = quick->next;
if(quick) quick = quick->next;
if(slow) slow = slow->next;
}
ListNode *second = reverseList(slow->next);
ListNode *first = head;
slow->next = NULL;
ListNode *tmp = dummy;
while(first) {
tmp->next = first;
tmp = tmp->next;
first = first->next;
if(second) {
tmp->next = second;
tmp = tmp->next;
second = second->next;
}
}
}
};