单链表 -- 合并两个链表后, 依然有序

原创已于 2023-07-09 16:34:19 修改 · 270 阅读

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#链表 #java #数据结构

于 2023-07-09 16:03:33 首次发布

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该文章描述了一个Java方法，用于合并两个已排序的链表并保持排序顺序。首先创建新链表，然后遍历每个链表，根据节点编号决定插入位置。如果有重复编号，会打印错误信息。最后提供了一个单元测试用例来验证功能的正确性。

一、思路:

1. 创建一个新链表让其指向其中一个链表, 如链表1

2. 遍历另一个链表(链表2), 在新链表中判断其应该插入的位置

3. 若到达新链表末尾, 但链表2依旧存在数据,其数据直接添加至新链表末尾

二、代码示例:

public static void mergeLinkedList(HeroNode head1, HeroNode head2) {
       if (Optional.ofNullable(head1.getNext()).isEmpty() || Optional.ofNullable(head2.getNext()).isEmpty()) {
            return;
       }

       // 创建一个新的链表
       HeroNode mergeLink = new HeroNode(0, "", "");
       // 将链表指向第一个链表中的数据
       mergeLink.setNext(head1.getNext());
       // 辅助
       HeroNode temp2 = head2.getNext();
       HeroNode temp = mergeLink;
       while (Optional.ofNullable(temp2).isPresent()) {
           HeroNode t = temp2.getNext();
           while (Optional.ofNullable(temp.getNext()).isPresent()) {
               if(temp.getNext().getNo() == temp2.getNo()) {
                    System.out.println("链表中编号不可重复!!!");
                    temp2 = t;
                    break;
               }
               if(temp.getNext().getNo() > temp2.getNo()) {
                   temp2.setNext(temp.getNext());
                   temp.setNext(temp2);
                   temp2 = t;
                   break;
               }
               temp = temp.getNext();
           }
           if (Optional.ofNullable(temp.getNext()).isEmpty() && temp.getNo() < temp2.getNo()) {
                temp.setNext(temp2);
                break;
            }
       }
       // 遍历输出
       HeroNode next = mergeLink.getNext();
       while (Optional.ofNullable(next).isPresent()) {
           System.out.println(next);
           next = next.getNext();
       }
}

三、单元测试:

@Test
public void test2() {
    HeroNode one = new HeroNode(1, "小刚", "笑面虎");
    HeroNode two = new HeroNode(3, "小红", "冷面教练");
    HeroNode three = new HeroNode(6, "小明", "笑里藏刀");
    HeroNode four = new HeroNode(8, "小李", "双刀剑客");
    HeroNode four2 = new HeroNode(9, "小李", "双刀剑客");
    HeroNode four3 = new HeroNode(10, "小李", "双刀剑客");

    SingleLinkedList singleLinkedList = new SingleLinkedList();

    singleLinkedList.addHeroMsgByOrder(one);
    singleLinkedList.addHeroMsgByOrder(two);
    singleLinkedList.addHeroMsgByOrder(three);
    singleLinkedList.addHeroMsgByOrder(four);
    singleLinkedList.addHeroMsgByOrder(four2);
    singleLinkedList.addHeroMsgByOrder(four3);

    HeroNode one1 = new HeroNode(2, "小刚", "笑面虎");
    HeroNode two1 = new HeroNode(4, "小红", "冷面教练");
    HeroNode three1 = new HeroNode(5, "小明", "笑里藏刀");
    HeroNode four1 = new HeroNode(7, "小李", "双刀剑客");

    SingleLinkedList singleLinkedList1 = new SingleLinkedList();

    singleLinkedList1.addHeroMsgByOrder(one1);
    singleLinkedList1.addHeroMsgByOrder(two1);
    singleLinkedList1.addHeroMsgByOrder(three1);
    singleLinkedList1.addHeroMsgByOrder(four1);

    SingleLinkedList.mergeLinkedList(singleLinkedList1.getHead(), singleLinkedList.getHead());
}