1.题目
https://leetcode.cn/problems/linked-list-cycle-ii/description/
2.解法
快慢指针:
(1)当全是环时,需要判断slow==head
(2)设头结点到第一次相遇点距离为m,入环节点到相遇点为k,快指针比慢指针多走m,这个m可以是绕环n圈。slow回到head,以相同的速度再走m-k则相遇为入环节点。
代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode* fast = head;
ListNode* slow = head;
while (fast != NULL && fast->next != NULL)
{
fast = fast->next->next;
slow = slow->next;
if (fast == slow)
{
if (slow == head) return slow;
slow = head;
break;
}
}
while (fast != NULL)
{
fast = fast->next;
slow = slow->next;
if (fast == slow)
{
return slow;
}
}
return NULL;
}
};