剑指Offer-56

来源：互联网发布：网络流行组合字编辑：程序博客网时间：2024/05/12 09:10
题目：

一个链表中包含环，如何找出环的入口结点？
实现

public class Solution56 {    public static Node findEntryNode(Node head){        if(head==null){            return null;        }        Node slow = head;        Node quick = head;        boolean flag = false;        while (quick != null && quick.next != null){            slow = slow.next;            quick = quick.next.next;            if(slow == quick){                flag = true;                System.out.println("有环");                break;            }        }        if(!flag){            System.out.println("无环");            return null;        }        slow = head;        while (slow!=quick){            slow=slow.next;            quick=quick.next;        }        return slow;    }    public static void main(String[] args) {        test01();        test02();        test03();    }    // 1->2->3->4->5->6    private static void test01() {        Node n1 = new Node(1);        Node n2 = new Node(2);        Node n3 = new Node(3);        Node n4 = new Node(4);        Node n5 = new Node(5);        Node n6 = new Node(6);        n1.next = n2;        n2.next = n3;        n3.next = n4;        n4.next = n5;        n5.next = n6;        Node result = findEntryNode(n1);        if (result != null) {            System.out.println(result.value);        }    }    // 1->2->3->4->5->6    //       ^        |    //       |        |    //       +--------+    private static void test02() {        Node n1 = new Node(1);        Node n2 = new Node(2);        Node n3 = new Node(3);        Node n4 = new Node(4);        Node n5 = new Node(5);        Node n6 = new Node(6);        n1.next = n2;        n2.next = n3;        n3.next = n4;        n4.next = n5;        n5.next = n6;        n6.next = n3;        Node result = findEntryNode(n1);        if (result != null) {            System.out.println(result.value);        }    }    // 1->2->3->4->5->6 <-+    //                |   |    //                +---+    private static void test03() {        Node n1 = new Node(1);        Node n2 = new Node(2);        Node n3 = new Node(3);        Node n4 = new Node(4);        Node n5 = new Node(5);        Node n6 = new Node(6);        n1.next = n2;        n2.next = n3;        n3.next = n4;        n4.next = n5;        n5.next = n6;        n6.next = n6;        Node result = findEntryNode(n1);        if (result != null) {            System.out.println(result.value);        }    }}
证明
阅读全文
0 0