[编程题] 合并两个有序链表--top100

合并两个有序链表

题目描述

示例

Java代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        //最终的生成的链表的头结点
        ListNode head = new ListNode(-1);
        ListNode reshead = head; //最终链表的哨兵
        ListNode l1head = l1;  //l1链表的哨兵
        ListNode l2head = l2;  ///l2链表的哨兵
        //当有一个链表到头的时候就退出
        while(!(l1head==null || l2head==null)){
            //当l1链表的值小于l2链表的值，就把它挂在最终的链表上
            if(l1head.val<=l2head.val){
                reshead.next = l1head;
                l1head = l1head.next;
                reshead = reshead.next;
            }else{//反之
                reshead.next = l2head;
                l2head = l2head.next;
                reshead = reshead.next;
            }
        }

        //判断剩余的是哪个链表没取完，直接挂在head链表后
        reshead.next = l1head==null ? l2head : l1head;

        //返回最终的链表
        return head.next;
    }
}

输出

标签：ListNode,val,编程,next,链表,l2,l1,top100
来源： https://www.cnblogs.com/jiyongjia/p/13252500.html