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69 lines (54 loc) · 2.24 KB
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class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def list_to_listnode(lst):
dummy = ListNode()
cur = dummy
for val in lst:
cur.next = ListNode(val)
cur = cur.next
return dummy.next
def listnode_to_list(node):
res = []
while node:
res.append(node.val)
node = node.next
return res
"""
2. Add Two Numbers / 两数相加 (Medium)
给你两个非空的链表,表示两个非负的整数。它们每位数字都是按照逆序的方式存储的,并且每个节点只能存储一位数字。
请你将两个数相加,并以相同形式返回一个表示和的链表。
你可以假设除了数字 0 之外,这两个数都不会以 0 开头。
You are given two non-empty linked lists representing two non-negative integers.
The digits are stored in reverse order, and each of their nodes contains a single digit.
Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Example 2:
Input: l1 = [0], l2 = [0]
Output: [0]
Example 3:
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
"""
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
resHead = resNode = ListNode()
tens = units = num = 0
while tens or l1 or l2:
num = (l1.val if l1 else 0) + (l2.val if l2 else 0) + tens
tens, units = num // 10, num % 10
resNode.next = ListNode(units)
resNode, l1, l2 = resNode.next, l1.next if l1 else None, l2.next if l2 else None
return resHead.next
if __name__ == "__main__":
solution = Solution()
print("Example 1:", listnode_to_list(solution.addTwoNumbers(
list_to_listnode([2, 4, 3]), list_to_listnode([5, 6, 4])))) # Expected: [7, 0, 8]
print("Example 2:", listnode_to_list(solution.addTwoNumbers(
list_to_listnode([0]), list_to_listnode([0])))) # Expected: [0]
print("Example 3:", listnode_to_list(solution.addTwoNumbers(
list_to_listnode([9, 9, 9, 9, 9, 9, 9]), list_to_listnode([9, 9, 9, 9])))) # Expected: [8,9,9,9,0,0,0,1]