环形链表的环长和入口求解

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环形链表的环长和入口求解

快慢双指针解法
求入口位置时，一个环形链表，我们假设链表的起始点到环的入口点的距离为L，环的周长为R，环的入口点到快慢指针的相遇位置的距离为X
快指针走的距离：F = L+X+nR
慢指针走的距离：S = L+X
因为快指针走的距离是慢指针的两倍，所以F = 2S。
这时：L+X+nR = 2 * (L + X)
L = nR - X

public class RollLinkedList {
    public static void main(String[] args) {
        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 = n1;
        System.out.println("环长"+findLoopLength(n1));
        System.out.println("入口" + findLoopEnter(n1));

    }

    public static int findLoopLength(Node head) {
        Node fast = head;
        Node slow = head;
        boolean loopExist = false;
        int len = 1;
        if (head == null) {
            return 0;
        }
        //判断是否有环
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                loopExist = true;
                System.out.println("链表有环");
                break;
            }
        }
        //求环长
        if (loopExist) {
            fast = fast.next;
            while (slow != fast) {
                fast = fast.next;
                len++;
            }
            return len;
        }
        //链表没有环
        return 0;
    }

    //环入口
    public static int findLoopEnter(Node head) {
        Node fast = head;
        Node slow = head;
        if (head == null) {
            return 0;
        }
        //判断是否有环
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                System.out.println("链表有环");
                //有环
                //记住相遇点的指针，一个从相遇点开始，一个从头开始，相遇的地方就是环的入口点
                Node tmp = head;
                //假设链表的开始就是入口点
                if (tmp == fast) {
                    return fast.value;
                }
                //如果入口不是在开始，则slow(temp）**从头**开始走l长度，
                //fast从相遇点继续（并没有从头开始走哦）向前走nr-x长度，
                //他们一定会在环入口处相遇
                while (true) {
                    fast = fast.next;
                    tmp = tmp.next;
                    if (fast == tmp) {
                        break;
                    }
                }
                return fast.value;

            }
        }
        return -1;
    }


}

class Node {
    public int value;
    public Node next;

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

参考https://blog.youkuaiyun.com/baidu_40931662/article/details/84306202#commentBox
https://blog.youkuaiyun.com/weixin_42146769/article/details/88430572