合并两个有序的链表

先自己实现一个单向的链表

package constxiong.interview;

/**
 * 单向链表
 * @author ConstXiong
 * @param <E>
 */
class SingleLinkedList<E> {
    
    int size = 0;
    
    Node<E> first;
    Node<E> last;
    
    public SingleLinkedList() {
        
    }
    
    public void add(E e) {
        Node<E> l = last;
        Node<E> item = new Node<E>(e, null);
        last = item;
        if (l == null) {
            this.first = item;
        } else {
            l.next = item;
        }
        size++;
    }
    /**
     * 打印链表
     * @param ll
     */
    public void print() {
        for (Node<E> item = first; item != null; item = item.next) {
            System.out.print(item + " ");
        }
    }

    /**
     * 单向链表中的节点
     * @author ConstXiong
     * @param <E>
     */
    public static class Node<E> {
        E item;
        Node<E> next;
        
        Node(E item, Node<E> next) {
            this.item = item;
            this.next = next;
        }
        

        public E get() {
            return item;
        }

        @Override
        public String toString() {
            return item.toString();
        }
        
    }
}

 

 

题目中链表是有序的,所以不需要考虑排序问题

mergeeSingleLinkedList 方法合并链表,思路

  • 获取两个链表中的首节点
  • 比较首节点大小,结果分别存入 small、large 节点
  • 把 small 节点存入新的链表,再比较获取 small.next 和 large,结果分别存入 small、large 节点
  • 直到 small.next 和 large 都为 null
package constxiong.interview;

import constxiong.interview.SingleLinkedList.Node;

/**
 * 链表两个有序列表
 * @author ConstXiong
 * @date 2019-11-06 09:37:14
 */
public class TestMergeLinkedList {

    public static void main(String[] args) {
        SingleLinkedList<Integer> ll1 = new SingleLinkedList<Integer>();
        ll1.add(3);
        ll1.add(8);
        ll1.add(19);
        
        SingleLinkedList<Integer> ll2 = new SingleLinkedList<Integer>();
        ll2.add(3);
        ll2.add(10);
        ll2.add(17);
        
        mergeeSingleLinkedList(ll1, ll2).print();
    }
    
    /**
     * 合并两个有序列表
     * @param ll1
     * @param ll2
     * @return
     */
    private static SingleLinkedList<Integer> mergeeSingleLinkedList(SingleLinkedList<Integer> ll1, SingleLinkedList<Integer> ll2) {
        if (isEmpty(ll1) || isEmpty(ll2)) {
            return isEmpty(ll1) ? ll2 : ll1;
        }
        SingleLinkedList<Integer> ll = new SingleLinkedList<Integer>();
        Node<Integer> ll1Node = ll1.first;
        Node<Integer> ll2Node = ll2.first;
        Node<Integer> small = ll1Node.get() <= ll2Node.get() ? ll1Node : ll2Node;
        Node<Integer> large = ll1Node.get() > ll2Node.get() ? ll1Node : ll2Node;
        do {
            ll.add(small.get());
            Node<Integer> smallNext = small.next;
            if (smallNext == null || large == null) {
                small = smallNext == null ? large : smallNext;
                large = null;
            } else {
                small = smallNext.get() <= large.get() ? smallNext : large;
                large = smallNext.get() > large.get() ? smallNext : large;
            }
        }
        while (small.next != null || large != null);
        return ll;
    }
    
    /**
     * 测试链表存储是否OK
     */
    public static void testSingleLinkedListIsOk() {
        SingleLinkedList<Integer> ll = new SingleLinkedList<Integer>();
        ll.add(3);
        ll.add(8);
        ll.add(19);
        ll.print();
    }
    
    
    private static boolean isEmpty(SingleLinkedList<Integer> ll) {
        if (ll == null || ll.size == 0) {
            return true;
        }
        return false;
    }
    
}

 

 

打印结果

3 3 8 10 17 


原文链接
 


 

 

posted @ 2020-01-06 09:29  ConstXiong  阅读(259)  评论(0编辑  收藏  举报