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);
}
}
