LeetCode-Merge Two Sorted Lists

Description：
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.

Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

题意：将所给的两个已排好序的链表，合成一个有序的链表

解法：因为两个链表都是已排好序的，所以我们可以从两个链表的开始结点逐一的进行比较，将值小的结点先添加到新的链表上；

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null) return l2;
        if(l2 == null) return l1;
        ListNode list = null;//链表的第一个结点
        if(l1.val < l2.val) {
            list = new ListNode(l1.val);
            l1 = l1.next;
        }
        else{
            list = new ListNode(l2.val);
            l2 = l2.next;
        }
        ListNode listNext = list;//链表的尾结点
        while(l1 != null && l2 != null){
            if(l1.val < l2.val){
                listNext.next = new ListNode(l1.val);
                l1 = l1.next;
                listNext = listNext.next;
            }
            else{
                listNext.next = new ListNode(l2.val);
                l2 = l2.next;
                listNext = listNext.next;
            }
        }
        while(l1 != null){
            listNext.next = new ListNode(l1.val);
            l1 = l1.next; 
            listNext = listNext.next;
        }//l1还有剩余结点
        while(l2 != null){
            listNext.next = new ListNode(l2.val);
            l2 = l2.next; 
            listNext = listNext.next;
        }//l2还有剩余结点
        return list;
    }
}