【Java】将两个有序链表合并为一个有序链表

朴灿烈づ我的快乐病毒、 2023-03-03 12:27 4阅读 0赞

public class Node
    {
        private Node next;
        private Integer number;
        Node(Integer number)
        {
            this.number=number;
            next=null;
        }
        Node()
        {
    
        }
        public Node getNext() {
            return next;
        }
        public void setNext(Node next) {
            this.next = next;
        }
        public Integer getNumber() {
            return number;
        }
        public void setNumber(Integer number) {
            this.number = number;
        }
    
    }

//合并链表
    public class SortForLink {
    
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            int[] arr1 = {1,3,6,8,23,34,56,77,90};
            int[] arr2 = {-90,34,55,79,87,98,123,234,567};
    
            Node link1 = new Node();
            Node link2 = new Node();
            Node link3 = new Node();
    
            getLink(arr1,link1);
            getLink(arr2,link2);
    
            mergeLink(link1, link2, link3);
    
            viewLink(link3);
    
    
    
        }
    
        /**
         * 构建链表
         * @param arr
         * @param link
         */
        public static void getLink(int[] arr, Node link) {
    
            for(int i = 0; i < arr.length; i++) {
                link.setNumber(arr[i]);
                link.setNext(new Node());
                link = link.getNext();
            }
    
        }
    
        /**
         * 输出链表
         * @param link
         */
        public static void viewLink(Node link) {
    
            while(link.getNumber() != null) {
                System.out.println(link.getNumber());
                link = link.getNext();
            }
        }
    
        /**
         * 合并链表
         * @param link1
         * @param link2
         * @param link3
         */
        public static void mergeLink(Node link1, Node link2, Node link3) {
    
            if( (link2.getNumber() != null) && (link1.getNumber() == null || link1.getNumber() >= link2.getNumber()) ) {
                link3.setNumber(link2.getNumber());
                link3.setNext(new Node());
                link2 = link2.getNext();
                link3 = link3.getNext();
                mergeLink(link1, link2, link3);
            }else if( (link1.getNumber() != null) && (link2.getNumber() == null || link2.getNumber() > link1.getNumber()) ) {
                link3.setNumber(link1.getNumber());
                link3.setNext(new Node());
                link1 = link1.getNext();
                link3 = link3.getNext();
                mergeLink(link1, link2, link3);
            }else {
                System.out.println("==没有了，结束啦！==");
            }
        }
    
    }

转载地址：[https://blog.csdn.net/qq\_43669007/article/details/91316984][https_blog.csdn.net_qq_43669007_article_details_91316984]

[https_blog.csdn.net_qq_43669007_article_details_91316984]: https://blog.csdn.net/qq_43669007/article/details/91316984