1. 问题描述
2. 解决办法:
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
dummy_head = ListNode(0)
current = dummy_head
while l1 and l2:
if l1.val <= l2.val:
current.next = ListNode(l1.val)
l1 = l1.next
else:
current.next = ListNode(l2.val)
l2 = l2.next
current = current.next
# while l1:
# current.next = ListNode(l1.val)
# l1 = l1.next
# current = current.next
# while l2:
# current.next = ListNode(l2.val)
# l2 = l2.next
# current = current.next
current.next = l1 or l2
return dummy_head.next
l2 = ListNode(5)
l22 = ListNode(6)
l23 = ListNode(4)
l2.next = l22
l22.next = l23
s = Solution()
a = s.mergeTwoLists(l2, l2)
print(a)
3. 个人记录:
1.这道题和求两个sortedList的中位数的那道题很像
2. 关于链表,几个注意的点:
要有一个变量不断记录当前节点
current.next=ListNode(l1.val)