判断一个单链表是否有环

用快慢指针法，慢指针一次走一步，快指针一次走两步，如果有环必定相遇

这里给出C++和Python 版

Python：

"""
Definition of ListNode
class ListNode(object):

    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""
class Solution:
    """
    @param head: The first node of the linked list.
    @return: True if it has a cycle, or false
    """
    def hasCycle(self, head):
        # write your code here
        if head is None:
            return False
        if head.next is None:
            return False
        slow = head.next
        fast = head.next
        if fast.next is not None:
            fast = fast.next
        while (slow.next is not None) and (fast.next is not None):
            if slow == fast:
                return True
            slow = slow.next
            fast = fast.next
            if fast.next != None:
                fast = fast.next
        return False

c++

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: True if it has a cycle, or false
     */
    bool hasCycle(ListNode *head) {
        // write your code here
        if (head == NULL)
            return false;
        if (head->next ==NULL||head->next->next==NULL)
            return false;
            ListNode *p = head ->next;
            ListNode *q = head ->next->next;
        while(p->next&&q->next) {
            if(p==q){
                return true;
            }
           p = p->next;
           q = q->next;
           if (q->next)
           q=q->next;
        }
        return false;
    }
};

时间复杂度O(N) 常数的空间复杂度