算法：判断单链表成环与否

最新推荐文章于 2024-04-20 12:42:04 发布

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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。通过创建两个指针，一个慢指针每次移动一步，另一个快指针每次移动两步，如果链表中有循环，这两个指针最终会相遇。

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

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

package com.bxh.algorithms.linklist;

/**
 * Created by bxh on 8/1/17.
 */

public class LinkListTest {
    public Node getNode(boolean isCircle, int circlePos) {
        Node circleNode = null;
        Node head = null;
        Node last = null;
        for (int i = 1; i < 6; i++) {
            if (head == null) {
                head = new Node(0);
                last = head;
            }
            Node node = new Node(i);
            if (i == circlePos) {
                circleNode = node;
            }
            if (last != null) {
                last.next = node;
            }
            last = node;
        }
        if (isCircle && last != null) {
            last.next = circleNode;
        }
        return head;
    }

    public boolean isCircle(Node head) {
        if (head == null)
            return false;
        Node n1 = head;//慢指针
        Node n2 = head;//快指针
        while (n2.next != null && n2.next.next != null) {
            n1 = n1.next;
            n2 = n2.next.next;
            if (n1 == n2)
                return true;
        }

        return false;
    }

    public static void test(){
        LinkListTest l = new LinkListTest();
        Node node1 = l.getNode(true,3);
        System.out.println("--true,3--"+l.isCircle(node1));
        Node node2 = l.getNode(false,3);
        System.out.println("--false,3--"+l.isCircle(node2));
    }
}