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输入两个递增的链表#xff0c;单个链表的长度为n#xff0c;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#xff1a;
{1,3,5}, {2,4,6}返回值#xff1a;
{1,2,3,4,5,6}原题地址#xff1a;https://www.nowcoder.com/practice/d8b6b4358f7742…
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 hereif (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 5ListNode 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 6ListNode 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);}
}
