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.

题目要求：

合并两个有序链表

注意：

不能开辟新的结点空间

解题思路：

1、归并排序，创建一个新的头结点，从头到尾分别遍历两个链表，并依次比较其大小关系，每次将头指针指向小的那个。

2、递归思想

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

// Merge combination method
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode head(0);
        ListNode *lst;
        lst=&head;
        while(l1 && l2){
            if(l1->val<=l2->val){
                lst->next=l1;
                l1=l1->next;
            }
            else{
                lst->next=l2;
                l2=l2->next;
            }
            lst=lst->next;
        }
        if(l1)
            lst->next=l1;
        if(l2)
            lst->next=l2;
        return head.next;
    }
};

// Recursive method
class Solution {
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(l2->next, l1);
            return l2;
        }
    }