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[LeetCode] 2. Add Two Numbers Problem Solving Report (Python)
Topic Address: https://leetcode.com/problems/add-two-numbers/
Title Description
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 contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
Solution one: first sum, and then build the list
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
num1 = ''
num2 = ''
while l1:
num1 = num1+str(l1.val)
l1 = l1.next
while l2:
num2 = num2+str(l2.val)
l2 = l2.next
add = str(int(num1[::-1])+int(num2[::-1]))[::-1]
l = ListNode(add[0])
tmp = l
for i in range(1, len(add)):
tmp.next = ListNode(add[i])
tmp = tmp.next
return l
Solution two: a carry using carry facilitate the completion of a traverse the outcome
Place two to note: If the list is not of equal length; if the list is still to complete the final sum carry bits.
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
ans = ListNode(0)
tmp = ans
carry = 0
while l1 and l2:
add = l1.val + l2.val + carry
tmp.next = ListNode(add % 10)
tmp = tmp.next
carry = int(add/10)
l1 = l1.next
l2 = l2.next
l = l1 if l1 else l2
while l:
add = l.val + carry
tmp.next = ListNode(add % 10)
tmp = tmp.next
carry = int(add / 10)
l = l.next
if carry:
tmp.next = ListNode(carry)
return ans.next
Solution three: recursive
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
if not l1 and not l2:
return
elif not (l1 and l2):
return l1 or l2
else:
if l1.val + l2.val < 10:
l3 = ListNode(l1.val + l2.val)
l3.next = self.addTwoNumbers(l1.next, l2.next)
else:
l3 = ListNode(l1.val + l2.val - 10)
l3.next = self.addTwoNumbers(l1.next, self.addTwoNumbers(l2.next, ListNode(1)))
return l3