合并有序两个单链表,合并后链表依然有序

Posted 2020-12-23 dong9012

最近在学习数据结构,特此记录一下,方便以后查阅.

 1 //定义一个类来管理我们的英雄 也就是链表
 2 class SingleLinkedList{
 3     //先初始化一个头节点,头节点不能动,用于寻找链表的头
 4     private HeroNode head = new HeroNode(0,"","");
 5 
 6     public HeroNode getHead() {
 7         return head;
 8     }
 9     
10 
11     public void addOrderByNo(HeroNode heroNode){
12         //定义一个变量来辅助遍历
13         HeroNode temp = head;
14         boolean flag = false;//用来判断数据书否存在的
15 
16         while (true){
17             //为链表的最后
18             if(temp.next==null){
19                 break;
20             }
21             //如果下一个节点的数字已经大于现在要添加的数了 证明已经找到了 当前要添加的位置
22             if(temp.next.no>heroNode.no){
23                 break;
24             }
25             //如果等于要添加的数 证明这个数据已经添加过了 不能在添加
26             if(temp.next.no==heroNode.no){
27                 flag = true;//证明存在
28                 break;
29             }
30             //没找到后移temp
31             temp = temp.next;
32         }
33 
34         //判断flag的值
35         if(!flag){
36             //插入数据到链表中
37             //1.先将添加的数据的next指向下一个数据
38             heroNode.next = temp.next;
39             //2.再将temp的next指向添加的数据
40             temp.next = heroNode;
41         }else{
42             System.out.printf("准备插入的英雄的编号 %d 已经存在了, 不能加入\n", heroNode.no);
43         }
44     }
45 }
46 
47 class HeroNode{
48     public int no;
49     public String name;
50     public String nickname;
51     public HeroNode next; //指向下一个节点
52 
53     public HeroNode(int no, String name, String nickname) {
54         this.no = no;
55         this.name = name;
56         this.nickname = nickname;
57     }
58 
59     @Override
60     public String toString() {
61         return "HeroNode{" +
62                 "no=" + no +
63                 ", name=‘" + name + ‘\‘‘ +
64                 ", nickname=‘" + nickname + ‘\‘‘ +
65                 ‘}‘;
66     }
67 }

 1 /**
 2      * 合并两个有序的单链表 使合并完成之后的新的链表依旧有序
 3      * @param hero1
 4      * @param hero2
 5      */
 6     public static HeroNode mergeList(HeroNode hero1, HeroNode hero2){
 7 
 8         //创建一个新的链表
 9         HeroNode hero3 = new HeroNode(0,"","");
10 
11         HeroNode temp1 = hero1.next;
12         HeroNode temp2 = hero2.next;
13         HeroNode temp3 = hero3;
14 
15         HeroNode next1 = null;//用于保存下一个节点
16         HeroNode next2 = null;
17         while(temp1!=null&&temp2!=null){
18             if(temp1.no<temp2.no){
19                 next1 = temp1.next;//先保存下一个节点
20                 temp3.next = temp1;//将较小的节点插入到新的节点后面
21                 temp3 = temp3.next;//将新的节点后移一位 注意:一定要后移
22                 temp1 = next1;//将节点后移一位
23             }else{
24                 next2 = temp2.next;
25                 temp3.next = temp2;
26                 temp3 = temp3.next;
27                 temp2 = next2;
28             }
29         }
30         if(temp1 == null){
31             while (temp2 !=null){
32                 next2 = temp2.next;
33                 temp3.next = temp2;
34                 temp3 = temp3.next;
35                 temp2 = next2;
36             }
37         }else{
38             while (temp1!=null){
39                 next1 = temp1.next;//先保存下一个节点
40                 temp3.next = temp1;//将较小的节点插入到新的节点后面
41                 temp3 = temp3.next;//将新的节点后移一位 注意:一定要后移
42                 temp1 = next1;//将节点后移一位
43             }
44         }
45         return hero3;
46     }

 1 public static void main(String[] args) {
 2         HeroNode heroNode1 = new HeroNode(1,"李1","木子李1");
 3         HeroNode heroNode2 = new HeroNode(2,"李2","木子李2");
 4         HeroNode heroNode3 = new HeroNode(3,"李3","木子李3");
 5         HeroNode heroNode4 = new HeroNode(4,"李4","木子李4");
 6 
 7         HeroNode heroNode5 = new HeroNode(5,"李5","木子李5");
 8         HeroNode heroNode6 = new HeroNode(6,"李6","木子李6");
 9         HeroNode heroNode7 = new HeroNode(7,"李7","木子李7");
10         HeroNode heroNode8 = new HeroNode(8,"李8","木子李8");
11 
12         SingleLinkedList list = new SingleLinkedList();
13 
14         list.addOrderByNo(heroNode1);
15         list.addOrderByNo(heroNode3);
16         list.addOrderByNo(heroNode5);
17         list.addOrderByNo(heroNode7);
18 
19         SingleLinkedList list2 = new SingleLinkedList();
20         list2.addOrderByNo(heroNode2);
21         list2.addOrderByNo(heroNode4);
22         list2.addOrderByNo(heroNode6);
23         list2.addOrderByNo(heroNode8);
24         
25         //合并两个链表
26         HeroNode heroNode = mergeList(list.getHead(), list2.getHead());
27     }