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(对于为改变链表结构的题目，一般也可以采用递归的方法)

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;}}

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（LeetCode 21）Merge Two Sorted Lists

题目要求：

注意：

解题思路：

代码：

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