LeetCode | Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */

 //思路是每次取两个list中的小者，归并排序的思想public class Solution {    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {        if (l1==null && l2==null) return null;        if (l1==null && l2!=null) return l2;        if (l1!=null && l2==null) return l1;                ListNode l1_node = l1;       //循环取两个list中的元素出来比较        ListNode l2_node = l2;       //相当于两个位置标志位，本list中的小者被取出后，标志位向后+1                ListNode new_head = new ListNode(0);  //相当于新的list的头，最终返回new_head.next        ListNode new_node = new_head;         //相当于新的list中的标志位，每次把比较出来的小者附到new_node的后边，同时new_node向后+1                while (l1_node!=null && l2_node!=null){            if(l1_node.val <= l2_node.val){                new_node.next = l1_node;          //取小者附在new_node的后面                l1_node = l1_node.next;           //标志node向后+1，略过这个小的node            }else{                new_node.next = l2_node;                l2_node = l2_node.next;            }            //至此，新的list：new_head->小者node->小node后边的list元素            new_node = new_node.next;   //第一次循环new_node从head变为最小的node，第二次循环变为第二小的node        }        if (l1_node == null) {new_node.next = l2_node;}    //说明l1不够，l2有剩余        if (l2_node == null) {new_node.next = l1_node;}    //说明l2不够，l1有剩余                return new_head.next;    }}