两个单链表相交的问题

单链表可能有环无环；

给定两个 单链表的头节点 head1和head2，这两个链表可能相交，也可能 不相交。请实现一个函数， 如果两个链表相交，请返回相交的 第一个节点；如果不相交，返回null 即可。

链表长度分别为 N M， 时间复杂度O（N + M）；额外空间 O（1） ；

package my_basic.class_3;

public class FindFirstIntersectNode {
    public static class Node{
        int value;
        Node next;
        public Node(int data) {
            this.value = data;
        }
    }
    /**
     * 找到相交节点
     * @return
     */
    public static Node getIntersectNode(Node head1,Node head2) {
        if (head1 == null || head2==null) {
            return null;
        }
        //环节点
        Node loop1 = getLoopNode(head1);
        Node loop2 = getLoopNode(head2);
        
        if (loop1 == null && loop2 == null) {
            return getNoLoop(head1,head2);
        }
        if (loop1!=null && loop2!=null) {
            return bothLoop(head1,loop1,head2,loop2);
        }
        return null;
        
    }
    
    //判断有没有环 如果有环 返回第一个入环节点
    //第1种：hashset实现 
    //第2种：一个快指针 一个慢指针 ； 如果相遇了， 快指针从头开始 每次一步走，首次相遇的节点就是第一个入环节点
    public static Node getLoopNode(Node head) {
        //2. 
        if (head == null || head.next == null || head.next.next == null ) {
            return null;
        }
        Node n1;  //slow
        Node n2;  //fast
        n1 = head.next;
        n2 = head.next.next;
        while(n1 != n2) {
            if (n2.next == null || n2.next.next == null ) {
                return null;
            }
            n1 = n1.next;
            n2 = n2.next.next;
        }
        n2 = head;
        while (n1 != n2) {
            n1 = n1.next;
            n2 = n2.next;
        }
        return n1;
        
        //1. 需要额外的哈希表做辅助空间
        /*HashSet<Object> set = new HashSet<>();
        Node n1 = head;  
        while (n1 != null) {
            if (set.contains(n1)) {
                return n1;
            }
            set.add(n1);
            n1 = n1.next;
        }
        return null;*/
    }
    
    //都没有环: 也可以借助hashset
    public static Node getNoLoop(Node head1, Node head2) {
        if (head1==null || head2 == null) {
            return null;
        }
        Node cur1 = head1;  //long list
        Node cur2 = head2;  //short  list
        int n=0;
        while (cur1.next != null) {
            n++;
            cur1 = cur1.next;
        }
        while (cur2.next != null) {
            n--;
            cur2 = cur2.next;
        }
        if (cur1 != cur2) {
            return null;
        } 
        cur1 = n>0 ? head1 : head2;
        cur2 = cur1 == head1 ? head2 : head1;
        
        n = Math.abs(n);
        while (n != 0) {
            n--;
            cur1 = cur1.next;
        }
        while (cur1 != cur2) {
            cur1 = cur1.next;
            cur2 = cur2.next;
        }
        return cur1;
    }
    
    //都有环: 也可以借助hashset
    public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
        if (loop1 == loop2 ) {
            Node cur1 = head1;  //long
            Node cur2 = head2;  //short
            int n=0;
            while (cur1.next != loop1) {
                n++;
                cur1 = cur1.next;
            }
            while (cur2.next != loop2) {
                n--;
                cur2 = cur2.next;
            }
            if (cur1 != cur2) {
                return null;
            } 
            cur1 = n>0 ? head1 : head2;
            cur2 = cur1 == head1 ? head2 : head1;
            
            n = Math.abs(n);
            while (n > 0) {
                n--;
                cur1 = cur1.next;
            }
            while (cur1 != cur2) {
                cur1 = cur1.next;
                cur2 = cur2.next;
            }
            return cur1;
        }else{
            Node n = loop1.next;
            while(n != loop1) {
                if (n == loop2) {
                    return loop1;
                }
                n = n.next;
            }
            return null;
        }
    }
    
    public static void main(String[] args) {
        // 1->2->3->4->5->6->7->null
        Node head1 = new Node(1);
//        head1.next = new Node(2);
//        head1.next.next = new Node(3);
//        head1.next.next.next = new Node(4);
//        head1.next.next.next.next = new Node(5);
//        head1.next.next.next.next.next = new Node(6);
//        head1.next.next.next.next.next.next = new Node(7);
//
//        // 0->9->8->6->7->null
        Node head2 = new Node(0);
//        head2.next = new Node(9);
//        head2.next.next = new Node(8);
//        head2.next.next.next = head1.next.next.next.next.next; // 8->6
//        System.out.println(getIntersectNode(head1, head2).value);
//
        // 1->2->3->4->5->6->7->4...
        head1 = new Node(1);
        head1.next = new Node(2);
        head1.next.next = new Node(3);
        head1.next.next.next = new Node(4);
        head1.next.next.next.next = new Node(5);
        head1.next.next.next.next.next = new Node(6);
        head1.next.next.next.next.next.next = new Node(7);
        head1.next.next.next.next.next.next.next = head1.next.next.next; // 7->4

//        // 0->9->8->2...
//        head2 = new Node(0);
//        head2.next = new Node(9);
//        head2.next.next = new Node(8);
//        head2.next.next.next = head1.next; // 8->2
//        System.out.println(getIntersectNode(head1, head2).value);

        // 0->9->8->6->4->5->6..
        head2 = new Node(0);
        head2.next = new Node(9);
        head2.next.next = new Node(8);
        head2.next.next.next = head1.next.next.next.next.next; // 8->6
        System.out.println(getIntersectNode(head1, head2).value);

    }
}