问题描述
解题思路:
1.利用快慢指针将链表分成两半
2.将后半段链表逆转(注意,将前半段最后一个指针赋值为null)
3.将两个链表拼接起来
即例如:
1->2->3->4
变成:
1->2
4->3
然后:
1->4->2->3
代码如下:
/**
* 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) return;
ListNode *slow = head,*fast = head;
while(fast->next && fast->next->next)
{
slow = slow->next;
fast = fast->next->next;
}
ListNode *mid = slow->next;
slow->next = NULL;
ListNode *pre = NULL;
while(mid)
{
ListNode *temp = mid->next;
mid->next = pre;
pre = mid;
mid = temp;
}
while(head && pre)
{
ListNode *next = head->next;
head->next = pre;
pre = pre->next;
head->next->next = next;
head = next;
}
}
};