lintcode:带环链表

带环链表

给定一个链表，判断它是否有环。

解题

定义两个指针p1 p2

p1每次向前走一步

p2每次向前走两步

当p2能赶上p1的时候说明有环

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @return: True if it has a cycle, or false
     */
    public boolean hasCycle(ListNode head) {  
        // write your code here
        if(head == null || head.next == null)
            return false;
        ListNode p1 = head;
        ListNode p2 = head;
        p2 = p2.next;
        while(p2.next!=null && p2.next.next!=null){
            if(p1 == p2){
                return true;
            }
            p1 = p1.next;
            p2 = p2.next.next;
        }
        return false;
    }
}

"""
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 == None or head.next == None:
            return False
        slow = head
        fast = head
        fast = fast.next
        while fast.next!=None and fast.next.next!=None:
            if fast == slow:
                return True
            else:
                slow = slow.next
                fast = fast.next.next
        return False