## Description

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


## Solutions

### 迭代实现

# Time: O(n)
# Space: O(1)
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = ListNode(0)
node = dummy
while l1 and l2:
if l1.val <= l2.val:
node.next = l1
l1 = l1.next
else:
node.next = l2
l2 = l2.next
node = node.next
if l1:
node.next = l1
if l2:
node.next = l2
# if l1 or l2:
#     node.next = l1 or l2
return dummy.next

# 208/208 cases passed (32 ms)
# Your runtime beats 85.62 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)


### 2. 递归方式

注意返回的时候要选什么。

# Time: O(n)
# Space: O(121)
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
if not l1 or not l2:
return l2 or l1
if l1.val <= l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2

# 208/208 cases passed (36 ms)
# Your runtime beats 62.62 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)


### 3. 原地迭代

# in-place, iteratively
def mergeTwoLists(self, l1, l2):
if None in (l1, l2):
return l1 or l2
dummy = cur = ListNode(0)
dummy.next = l1
while l1 and l2:
if l1.val < l2.val:
l1 = l1.next
else:
nxt = cur.next
cur.next = l2
tmp = l2.next
l2.next = nxt
l2 = tmp
cur = cur.next
cur.next = l1 or l2
return dummy.next