[LeetCode]Linked List Cycle

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本文介绍了一种使用快慢指针的方法来判断链表中是否存在循环。通过一个快速指针（每次移动两步）和一个慢速指针（每次移动一步），如果链表存在循环，则这两个指针最终会相遇。

Given a linked list, determine if it has a cycle in it.

Follow up:

Can you solve it without using extra space?

Analysis:

two pointers.
One pointer is slow (1 step a time)
One pointer is fast (2 steps a time)
If there is a cycle, the two pointers will eventually meet (equal).

java

public boolean hasCycle(ListNode head){
		if(head == null || head.next == null) return false;
		ListNode p1 = head;
		ListNode p2 = head;
		while(p1!=null){
			p1 = p1.next;
			if(p2.next!=null)
				p2 = p2.next.next;
			else 
				return false;
			if(p1 == p2)
				return true;
		}
		return false;
	}

c++

bool hasCycle(ListNode *head) {
    if(head == NULL)
        return false;
    ListNode *slow = head;
    ListNode *fast = head->next;
    while(slow !=NULL && fast !=NULL){
        
        if(slow == fast)
            return true;
        slow = slow->next;
        fast = fast->next;
        if(fast ==NULL)
            return false;
        fast = fast->next;
    }
    return false;
    }