[LeetCode]-02.-Add-two-Numbers
Problem
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.
[해석] 주어진 linked list 2개의 각각의 수를 reverse해서 더한 값을 return 하시오.
example1:
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Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
example2:
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Input: l1 = [0], l2 = [0]
Output: [0]
example3:
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Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
constraints(조건):
- The number of nodes in each linked list is in the range
[1, 100]. 0 <= Node.val <= 9- It is guaranteed that the list represents a number that does not have leading zeros.
[Solution]
이 문제를 풀기 위해 떠오른 방식
- 숫자를 거꾸로 정렬해서 합치기
- 계산!
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# Definition for singly-linked list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
num1 = ''
num2 = ''
for n in reversed(l1):
num1 += str(n)
for m in reversed(l2):
num2 += str(m)
result = list(str(int(num1) + int(num2)))
answer = ListNode(result[0], None)
→ 이 방법 사용시 .. listNode는 reversible 하지않다는 errorcode가 뜸
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class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
num = ''
num2 = ''
while l1:
num1 = str(now.val) + num1
now = now.next
while l2:
num2 = str(now.val) + num2
now = now.next
result = list(str(int(num1) + int(num2)))
answer = ListNode*result[0], None)
for i in range(1, len(result)):
temp = ListNode(result[i], answer)
answer = temp
return answer