# 0021. Merge Two Sorted Lists
###### tags: `Leetcode` `Medium` `Linked List`
Link: https://leetcode.com/problems/merge-two-sorted-lists/
## 思路
### 思路一
递归
### 思路二
当 l1 和 l2 都不是空链表时,判断 l1 和 l2 哪一个链表的头节点的值更小,将较小值的节点添加到结果里,当一个节点被添加到结果里之后,将对应链表中的节点向后移一位。
## Code
### 思路一
```c=
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 == nullptr){
return l2;
}
else if(l2 == nullptr){
return l1;
}
else if(l1->val < l2->val){
l1->next = mergeTwoLists(l1->next, l2);
return l1;
}
else{
l2->next = mergeTwoLists(l1, l2->next);
return l2;
}
}
};
```
### 思路二
```c=
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 == nullptr){
return l2;
}
if(l2 == nullptr){
return l1;
}
ListNode* head = new ListNode;
ListNode* prev = head;
while(l1!=nullptr && l2!=nullptr){
if(l1->val < l2->val){
prev->next = l1;
l1 = l1->next;
}
else{
prev->next = l2;
l2 = l2->next;
}
prev = prev->next;
}
prev->next = l1 == nullptr?l2:l1;
return head->next;
}
};
```
## Result
### 思路一
### 思路二
Runtime: 8 ms, faster than **70.36%** of C++ online submissions for Merge Two Sorted Lists.
Memory Usage: 14.8 MB, less than **73.95%** of C++ online submissions for Merge Two Sorted Lists.