两个单链表相交的一系列问题

Posted 2020-12-27 星辰之衍

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；如果不想交，返回null即可。
要求：如果链表1的长度为N，链表2的长度为M，时间复杂度请达到O(N+M)，额外空间复杂度请达到O(1)。

package chj;

public class Problem_11_FindFirstIntersectNode {


    public static class Node {
        public int value;
        public Node next;


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


    public static Node getIntersectNode(Node head1, Node head2) {
        if (head1 == null || head2 == null) {
            return null;
        }
        Node loop1 = getLoopNode(head1);
        Node loop2 = getLoopNode(head2);
        if (loop1 == null && loop2 == null) {
            return noLoop(head1, head2);
        }
        if (loop1 != null && loop2 != null) {
            return bothLoop(head1, loop1, head2, loop2);
        }
        return null;
    }


    public static Node getLoopNode(Node head) {
        if (head == null || head.next == null || head.next.next == null) {
            return null;
        }
        Node n1 = head.next; // n1 -> slow
        Node n2 = head.next.next; // n2 -> fast
        while (n1 != n2) {
            if (n2.next == null || n2.next.next == null) {
                return null;
            }
            n2 = n2.next.next;
            n1 = n1.next;
        }
        n2 = head; // n2 -> walk again from head
        while (n1 != n2) {
            n1 = n1.next;
            n2 = n2.next;
        }
        return n1;  //再次相交时(n1==n2),第一个入环的节点返回
    }


    public static Node noLoop(Node head1, Node head2) {
        if (head1 == null || head2 == null) {
            return null;
        }
        Node cur1 = head1;
        Node cur2 = head2;
        int n = 0;
        while (cur1.next != null) {
            n++;
            cur1 = cur1.next;
            //System.out.println("n===="+n);
            //System.out.println("cur1.value="+cur1.value);
        }
        System.out.println("链表1遍历完时，cur1指针对应的value值:"+cur1.value);
        while (cur2.next != null) {
            n--;
            cur2 = cur2.next;
            //System.out.println("n===="+n);
            //System.out.println("cur2.value="+cur2.value);
        }
        System.out.println("链表2遍历完时，cur2指针对应的value值:"+cur2.value);
        
        if (cur1 != cur2) {
            return null;
        }
        cur1 = n > 0 ? head1 : head2;
        cur2 = cur1 == head1 ? head2 : head1;
        //System.out.println("n的值 = "+n);
        n = Math.abs(n);
        while (n != 0) {
            n--;
            cur1 = cur1.next;
        }
        while (cur1 != cur2) {
            cur1 = cur1.next;
            cur2 = cur2.next;
        }
        return cur1;
    }


    public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
        Node cur1 = null;
        Node cur2 = null;
        if (loop1 == loop2) {
            cur1 = head1;
            cur2 = head2;
            int n = 0;
            while (cur1 != loop1) {
                n++;
                cur1 = cur1.next;
            }
            System.out.println("链表1遍历完 时,cur1指针对应的value值:"+cur1.value);
            while (cur2 != loop2) {
                n--;
                cur2 = cur2.next;
            }
            cur1 = n > 0 ? head1 : head2;
            cur2 = cur1 == head1 ? head2 : head1;
            System.out.println("n = "+n);  // n = len_head1-len_head2 = 7-9 = -2
            n = Math.abs(n);
            while (n != 0) {
                n--;
                cur1 = cur1.next;
            }
            while (cur1 != cur2) {
                cur1 = cur1.next;
                cur2 = cur2.next;
            }
            return cur1;
        } else {
            cur1 = loop1.next;
            while (cur1 != loop1) {
                if (cur1 == loop2) {
                    return loop1;
                }
                cur1 = cur1.next;
            }
            return null;
        }
    }


    public static void main(String[] args) {
        // 1->2->3->4->5->6->8->12->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(8);
        head1.next.next.next.next.next.next.next = new Node(12);


        // 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("Common IntersectNode: " + getIntersectNode(head1, head2).value);
        //System.out.println(head2.next.next.next.next.value);
        System.out.println("==========分割线==========");

//      // 1->2->3->4->5->6->7->4...  head1为有环链表，节点为4
        head1 = new Node(1);  //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);
        head1.next.next.next.next.next.next = head1.next.next.next; // 7->4
//
//
//      // 0->9->8->2...   head2为有环节点，节点为4(其链表组成为：0->9->8->2->3->4->5->6->7->4... )
        head2 = new Node(0);   //Node head1 =new Node(1)
        head2.next = new Node(9);
        head2.next.next = new Node(8);
        head2.next.next.next = head1.next; // 8->2
        System.out.println("head2.next.next.next.next.next的value值:"+head2.next.next.next.next.next.value);
        System.out.println("--IntersectNode--: "+ getIntersectNode(head1, head2).value);


        // 0->9->8->6->4->5->6.. ?      head2为有环节点
        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("==IntersectNode==: "+ getIntersectNode(head1, head2).value);
        System.out.println("head2.next.next.next.next.next的value值为："+head2.next.next.next.next.next.value);

    }


}

输出：

链表1遍历完时，cur1指针对应的value值:12
链表2遍历完时，cur2指针对应的value值:12
Common IntersectNode: 6
==========分割线==========
head2.next.next.next.next.next的value值:4
链表1遍历完 时,cur1指针对应的value值:4
n = -2
--IntersectNode--: 2
==IntersectNode==: 4
head2.next.next.next.next.next的value值为：5