问题描述：

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

分析：

初步想法：

构建一个新的链表，遍历比较L1和L2上的节点，小的加入新的链，然后指向下一个节点，直到有一条链的节点为空，把剩余的链的部分接到新的链上。

代码如下：

  public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null) return l2;
        if(l2 == null) return l1;
        ListNode first = new ListNode(0);
        ListNode res = first;
        while(l1 != null && l2 != null){
            if(l1.val <= l2.val) {
                res.next = new ListNode(l1.val);
                l1 = l1.next;    
            }else{
                res.next = new ListNode(l2.val);
                l2 = l2.next;
            }
            res = res.next;   
        }
        if(l1 == null) res.next = l2;
        else res.next = l1;
        return first.next;
    }

方法二：（递归）

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null) return l2;
        if(l2 == null) return l1;
        if(l1.val < l2.val){
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        }else{
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }