判断一链表是否有环，求环的第一个节点和环的长度

第一种方法：直接遍历时，用hashset存放，判断是否存在环

第二种方法：使用快慢指针

public class CycleLinkedList {

    public static void main(String[] args) {

        Node head = new Node(1);
        Node node3 = new Node(3);
        head.next = node3;
        head.next.next = new Node(5);
        head.next.next.next = new Node(7);
        head.next.next.next.next = new Node(9);
        head.next.next.next.next.next = node3;
        System.out.println("是否有环：" + hasCycle(head));
        Node enterNode = getEnterNode(head);
        System.out.println("环的入口：" + enterNode.value);
        System.out.println(getCycleSize(enterNode));

    }

    // 环的长度
    public static Integer getCycleSize(Node node) {
        Node start = node;
        int len = 1;
        while (node.next != start) {
            len++;
            node = node.next;
        }
        return len;
    }

    // 如果有环的话，慢指针和快指针一定会在环上的某个点相遇，但不一定是环的入口
    public static boolean hasCycle(Node head) {
        Node fast = head;
        Node slow = head;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (slow == fast) {
                return true;
            }
        }
        return false;
    }

    // 环的入口节点
    public static Node getEnterNode(Node head) {
        // 先求出快慢指针的相遇点
        Node fast = head;
        Node slow = head;
        Node cross = null;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (slow == fast) {
                cross = fast;
                break;
            }
        }
        // 从链表头部和相遇点开始，每次移动一个节点,他们相遇点就是环的入口
        Node start = head;
        while (start != null && cross != null) {
            if (start == cross) {
                return cross;
            }
            start = start.next;
            cross = cross.next;
        }
        return null;
    }

    public static class Node {

        Node next;
        int value;

        public Node(int value) {
            super();
            this.value = value;
        }
    }
}

 

 

来源：https://www.cnblogs.com/moris5013/p/11640652.html