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# 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

# 来源:力扣(LeetCode)
# 链接:https: // leetcode-cn.com/problems/merge-two-sorted-lists

# Definition for singly-linked list.

class ListNode:
def __init__(self, x):
self.val = x
self.next = None

class Solution:
def mergeTwoLists(self, l1, l2):
prehead = ListNode(-1)

prev = prehead
while l1 and l2:
if l1.val <= l2.val:
prev.next = l1
l1 = l1.next
else:
prev.next = l2
l2 = l2.next
prev = prev.next
prev.next = l1 if l1 is not None else l2

return prehead.next

def __init__(self, arr1, arr2):
n1 = ListNode(arr1[0])
x1 = ListNode(arr2[0])
self.mergeTwoLists(n1, x1)
print(self.mergeTwoLists(n1, x1).val)

Solution([1, 2, 4], [1, 3, 4])

通过比较两个列表节点的大小,替换节点的方式,拼接出一个由小到大排序的有序单向列表,这题对于我来说,难点在于列表的实现,由于之前没怎么接触过链表,还得慢慢去理解。