Leetcode Linked List Cycle

标签：

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

Leetcode Linked List Cycle

标签：

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