2. Add Two Numbers
难度: 中等
刷题内容
原题连接
- https://leetcode.com/problems/add-two-numbers
内容描述
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.
解题方案
思路 1
全部变成数字做加法再换回去呗，这多暴力，爽！
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
val1, val2 = [l1.val], [l2.val]
while l1.next:
val1.append(l1.next.val)
l1 = l1.next
while l2.next:
val2.append(l2.next.val)
l2 = l2.next
num1 = ''.join([str(i) for i in val1[::-1]])
num2 = ''.join([str(i) for i in val2[::-1]])
tmp = str(int(num1) + int(num2))[::-1]
res = ListNode(tmp[0])
run_res = res
for i in range(1, len(tmp)):
run_res.next = ListNode(tmp[i])
run_res = run_res.next
return res
思路 2
可以使用递归，每次算一位的相加
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
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)
tmp = ListNode(1)
tmp.next = None
l3.next = self.addTwoNumbers(l1.next, self.addTwoNumbers(l2.next ,tmp))
return l3
