Leetcode 链表 Linked List Cycle II

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Linked List Cycle II

Total Accepted: 20444 Total
Submissions: 66195My Submissions

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Follow up:

Can you solve it without using extra space?

题意：给定一个单链表，判断该链表中是否存在环，如果存在，返回环开始的节点

思路：

1.定义两个指针，快指针fast每次走两步，慢指针s每次走一次，如果它们在非尾结点处相遇，则说明存在环

2.若存在环，设环的周长为r，相遇时，慢指针走了 slow步，快指针走了 2s步，快指针在环内已经走了 n环，

则有等式 2s = s + nr => s = nr

3.在相遇的时候，另设一个每次走一步的慢指针slow2从链表开头往前走。因为 s = nr，所以两个慢指针会在环的开始点相遇

复杂度：时间O(n)

struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

ListNode *detectCycle(ListNode *head) {
	if(!head || !head->next) return NULL;
	ListNode *fast, *slow, *slow2;
	fast = slow = slow2 = head;
	while(fast && fast->next){
		fast = fast->next->next;
		slow = slow->next;
		if(fast == slow && fast != NULL){
			while(slow->next){
				if(slow == slow2){
					return slow;
				}
				slow = slow->next;
				slow2 = slow2->next;
			}
		}
	}
	return NULL;
}