Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
题目要求不修改链表结构,同时不用额外的内存,去找到链表中环的起点。
我们可以通过快慢指针很容易判断有没有环,接下来,再把快指针的步伐也变为1并从head开始走,那么快慢指针再次相遇对应的指针就是环的起点。
解释:可以把问题抽象成一个直线和圈的结构,
粉色点表示相遇点,那么快指针走过a+b+c+b,慢指针走过a+b. 又2(a+b)=a+b+c+b,所以c=a.
/**
* 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;
if (head == nullptr) return head;
while (fast->next != nullptr && fast->next->next != nullptr) {
fast = fast->next->next;
slow = slow->next;
if (fast == slow) break;
}
if (fast->next == nullptr || fast->next->next == nullptr) return nullptr;
fast = head;
while (fast != nullptr && slow != nullptr) {
if (fast == slow) return fast;
fast = fast->next;
slow = slow->next;
}
return nullptr;
}
};