LeetCode-21. Merge Two Sorted Lists

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本文介绍了一种将两个已排序的链表合并为一个新排序链表的方法。提供了两种实现方式：迭代法和递归法，并附带了详细的Python代码示例。通过比较节点值，选择较小者加入新链表，最终返回合并后的链表。

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题目：

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

solution from discuss：

这是一个链表问题，有迭代和递归方法：

# iteratively
def mergeTwoLists1(self, l1, l2):
dummy = cur = ListNode(0)#注意是（）而不是[]
while l1 and l2:
if l1.val < l2.val:
cur.next = l1
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur = cur.next
cur.next = l1 or l2    #.next
return dummy.next

# recursively
def mergeTwoLists2(self, l1, l2):
if not l1 or not l2:
return l1 or l2
if l1.val < l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2

# 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
'''