# 21.Merge Two Sorted Lists

**21.Merge Two Sorted Lists**

难度:Easy

> 将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

```
示例：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4
```

简单递归：

```
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(!l1) return l2;
        if(!l2) return l1;
        ListNode* root; 
         if(l1->val < l2->val )
            {
                root=l1;
             l1=l1->next;
            }
            else 
                {
                root=l2;
                l2=l2->next;
            }
        root->next=mergeTwoLists(l1,l2);
        return root;
       
    }
};
```

> 执行用时 : 16 ms, 在Merge Two Sorted Lists的C++提交中击败了96.90% 的用户\
> 内存消耗 : 8.8 MB, 在Merge Two Sorted Lists的C++提交中击败了93.70% 的用户


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