原题:
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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/merge-two-sorted-lists
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解法一:(递归法)
在我看来,递归的思路是非常简单的,我们只需要将递归函数当成已经处理完毕的一个引用变量,由于最后递归函数是会递归到最简单的情况的之后在递归回去,所以我们只需要处理python (l1 == None or l2 == None)
这种情况其他复杂情况直接往下继续递归。
程序如下:
程序一:
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if l1 == None or l2==None:
if l1 != None:
return l1
elif l2 != None:
return l2
else:
return None
if l1.val > l2.val:
l2.next = self.mergeTwoLists(l1,l2.next)
return l2
else:
l1.next = self.mergeTwoLists(l1.next,l2)
return l1
程序二:
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if l1 and l2:
if l1.val > l2.val: l1, l2 = l2, l1
l1.next = self.mergeTwoLists(l1.next, l2)
return l1 or l2
解法二:(迭代)
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
pro = ans = ListNode(0)
if l1 == None or l2==None:
if l1 != None:
return l1
elif l2 != None:
return l2
else:
return None
while l1 != None and l2 != None:
if l1.val > l2.val:
ans.next = l2
l2 = l2.next
ans = ans.next
else:
ans.next = l1
l1 = l1.next
ans = ans.next
if l1 == None or l2==None:
if l1 != None:
ans.next = l1
elif l2 != None:
ans.next = l2
return pro.next