Java面试题：链表-合并两个排序的链表

描述

输入两个递增的链表，单个链表的长度为n，合并这两个链表并使新链表中的节点仍然是递增排序的。

示例

输入：
{1,3,5}, {2,4,6}

返回值：
{1,2,3,4,5,6}

原题地址：https://www.nowcoder.com/practice/d8b6b4358f774294a89de2a6ac4d9337

代码实现

package com.example.demo.linked;


public class ListNode {
    int val;
    ListNode next = null;

    public ListNode(int val) {
        this.val = val;
    }
}

package com.example.demo.linked;

public class LinkUtil {
    public static void printNodeList(ListNode head) {
        ListNode current = head;

        while (current != null) {
            System.out.print(current.val + " ");
            current = current.next;
        }
        
        System.out.println();
    }
}

package com.example.demo.linked;

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * @param pHead1 ListNode类
     * @param pHead2 ListNode类
     * @return ListNode类
     */
    public ListNode Merge(ListNode pHead1, ListNode pHead2) {
        // write code here

        if (pHead1 == null) {
            return pHead2;
        } else if (pHead2 == null) {
            return pHead1;
        }

        ListNode head = new ListNode(0);
        ListNode current = head;

        while (true) {
            if (pHead1 == null) {
                current.next = pHead2;
                break;
            } else if (pHead2 == null) {
                current.next = pHead1;
                break;
            } else {
                if (pHead1.val <= pHead2.val) {
                    current.next = pHead1;
                    pHead1 = pHead1.next;
                } else {
                    current.next = pHead2;
                    pHead2 = pHead2.next;
                }

                current = current.next;
            }
        }

        return head.next;
    }

    public static void main(String[] args) {
        // 1 3 5
        ListNode listNode1 = new ListNode(1);
        ListNode listNode2 = new ListNode(3);
        ListNode listNode3 = new ListNode(5);

        listNode1.next = listNode2;
        listNode2.next = listNode3;
        LinkUtil.printNodeList(listNode1);

        // 2 4 6
        ListNode listNodeA = new ListNode(2);
        ListNode listNodeB = new ListNode(4);
        ListNode listNodeC = new ListNode(6);

        listNodeA.next = listNodeB;
        listNodeB.next = listNodeC;
        LinkUtil.printNodeList(listNodeA);

        // 合并链表
        ListNode listNode = new Solution().Merge(listNode1, listNodeA);

        LinkUtil.printNodeList(listNode);

    }
}