每日一题 为了工作 2020 03013 第十一题

/**
* 问题：
* 打印两个有序链表的公共部分
* 给定两个有序链表的头指针head1和head2，打印出这两个链表的公共部分
* 分析：
* 因为是有序链表，所以从两个链表的头开始进行如下判断
* 1.如果head1的值小于head2的值，则head1往下移动。
* 2.如果head1的值大于head2的值，则head2往下移动。
* 3.如果head1的值等于head2的值，则打印出这个值，然后head1和head2都往下移动。
* 4.head1和head2如果有一个移动到null，则程序结束。
* @author 雪瞳
*
*/

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

public class CommandPart {
    
    public void printCommandPart(Node head1, Node head2) {
        System.out.println("command part is ...");
        while(head1!=null && head2!=null) {
            if(head1.value == head2.value) {
                System.out.println(head1.value);
                head1 = head1.next;
                head2 = head2.next;
            }else if(head1.value > head2.value) {
                head2 = head2.next;
            }else if(head1.value < head2.value) {
                head1 = head1.next;
            }
        }
    }
}

public class tetsGetCommanPart {
    public static void main(String[] args) {
        CommandPart test = new CommandPart();
        tetsGetCommanPart show = new tetsGetCommanPart();
        //准备数据
        Node head1 = new Node(3);
        Node head2 = new Node(5);
        
        Node n1 = new Node(11);
        Node n2 = new Node(22);
        Node n3 = new Node(33);
        Node n4 = new Node(44);
        Node n5 = new Node(55);
        Node n6 = new Node(66);
        Node n7 = new Node(77);
        Node n8 = new Node(88);
        Node n9 = new Node(99);
        
        //连接链表
        head1.next=n1;
        n1.next=n3;
        n3.next=n4;
        n4.next=n5;
        n5.next=n7;
        n7.next=n9;
        
        head2.next=n3;
        n3.next=n4;
        n4.next=n5;
        n5.next=n8;
        n8.next=n9;
        
        show.showLink(head1);
        show.showLink(head2);
        
        test.printCommandPart(head1, head2);
    }
    public void showLink(Node no) {
        while(no!=null) {
            System.out.print(no.value+"	");
            no=no.next;
        }
        System.out.println();
    }
}

*运行结果