找出两个无环单链表的相交点

题目

如图所示，找出两个单链表的相交点，若没有返回null

最好不要使用其他的辅助的数据结构，降低空间复杂度

代码

class Node {
    Node next;
}

public class Main {
    public static void main(String[] args) {
        Node head1 = new Node();
        head1.next = new Node();
        head1.next.next = new Node();
        head1.next.next.next = new Node();
        // 打印设置的相交点
        System.out.println("设置相交点"+head1.next.next);
        // 下个链表
        Node head2 = new Node();
        head2.next = head1.next.next;
        // 下个链表
        Node head3 = new Node();
        head3.next = new Node();
        head3.next.next = new Node();
        head3.next.next.next = new Node();
        // 结果
        Node res = findCrossNode(head1, head2);
        System.out.println("获取相交点"+res);
        res = findCrossNode(head1, head3);
        System.out.println("获取相交点"+res);
    }

    private static Node findCrossNode(Node head1, Node head2) {
        // 1. 判断是否相交
        Node n1 = head1;
        Node n2 = head2;
        int len1 = 1;
        int len2 = 1;
        while (n1.next != null) {
            n1 = n1.next;
            len1++;
        }
        while (n2.next != null) {
            n2 = n2.next;
            len2++;
        }
        if (n1 != n2) {
            return null;
        } else { // 到这里确定是相交的
            Node longList = head1;
            Node shortList = head2;
            if (len1 < len2) {
                longList = head2;
                shortList = head1;
            }
            int dis = Math.abs(len1-len2);
            // 长的先走几步
            for (int i = 0; i < dis; i++) {
                longList = longList.next;
            }
            while (longList != null && shortList != null) {
                if (longList == shortList){
                    return longList;
                }
                longList = longList.next;
                shortList = shortList.next;
            }
        }
        return null;
    }
}