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  • Leetcode Linked List Cycle

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

    Follow up:
    Can you solve it without using extra space?


    解题思路:

    If we have 2 pointers - fast and slow. It is guaranteed that the fast one will meet the slow one if there exists a circle.

    The problem can be demonstrated in the following diagram:

    复杂度O(n)的方法,使用两个指针slow,fast。两个指针都从表头开始走,slow每次走一步,fast每次走两步,如果fast遇到null,则说明没有环,返回false;如果slow==fast,说明有环,并且此时fast超了slow一圈,返回true。

    为什么有环的情况下二者一定会相遇呢?因为fast先进入环,在slow进入之后,如果把slow看作在前面,fast在后面每次循环都向slow靠近1,所以一定会相遇,而不会出现fast直接跳过slow的情况。


    Java code

    /**
     * Definition for singly-linked list.
     * class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) {
     *         val = x;
     *         next = null;
     *     }
     * }
     */
    public class Solution {
        public boolean hasCycle(ListNode head) {
            ListNode fast = head; //fast pointer moves two steps forward
            ListNode slow = head; //slow pointer moves one step forward
          
            while(fast != null && fast.next != null){
                slow = slow.next;
                fast = fast.next.next;
                if(slow == fast){
                    return true;
                }
            }
            return false;
        }
    }

    Reference:

    1. http://www.programcreek.com/2012/12/leetcode-linked-list-cycle/

    2. http://www.cnblogs.com/hiddenfox/p/3408931.html

    3. http://yucoding.blogspot.com/2013/10/leetcode-question-linked-list-cycle.html

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  • 原文地址:https://www.cnblogs.com/anne-vista/p/4960786.html
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