LeetCode 2 Add Two Numbers
题目-两数相加
【英文版】https://leetcode.com/problems/add-two-numbers/
【中文版】https://leetcode-cn.com/problems/add-two-numbers/
给出两个非空的链表用来表示两个非负的整数。其中，它们各自的位数是按照逆序的方式存储的，并且它们的每个节点只能存储一位数字。
如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外，这两个数都不会以 0 开头。
示例
输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
暴力求解
先求出输入链表l1,l2的和(traverse(self,l)),再用取余除10的方法逐个取出和的位数放入链表。
★ 时间复杂度：\(O(max(m,n))\)
★ 空间复杂的：\(O(max(m,n))+1\)
Brute Force -python-1
class Solution(object):
def traverse(self,l):
i=1
sum=0
while l is not None:
sum+=l.val*i
i=i*10
l=l.next
return sum
def addTwoNumbers(self, l1, l2):
l1_sum=self.traverse(l1)
l2_sum=self.traverse(l2)
sum=l1_sum+l2_sum
remainder = sum % 10
result=ListNode(remainder)
sum = int(sum/10)
pointer=result
while sum>0:
remainder=sum%10
pointer.next=ListNode(remainder)
pointer=pointer.next
sum =int(sum/10)
return result
执行用时 :64 ms, 在所有 Python 提交中击败了73.10%的用户
内存消耗 :11.8 MB, 在所有 Python 提交中击败了27.89%的用户
代码可精简部分
1. 将返回链表初始化为哑结点，否则要写重复的语句要初始化表头的值
Brute Force -python-2
class Solution(object):
def traverse(self,l):
i=1
sum=0
while l is not None:
sum+=l.val*i
i=i*10
l=l.next
return sum
def addTwoNumbers(self, l1, l2):
l1_sum=self.traverse(l1)
l2_sum=self.traverse(l2)
sum=l1_sum+l2_sum
if l1 is not None or l2 is not None:
if sum==0:
return ListNode(0)
dummyHead=ListNode(0)
pointer=dummyHead
while sum>0:
remainder=sum%10
pointer.next=ListNode(remainder)
pointer=pointer.next
sum =int(sum/10)
return dummyHead.next
Runtime: 60 ms, faster than 41.80% of Python online submissions for Add Two Numbers.
Memory Usage: 11.8 MB, less than 60.29% of Python online submissions for Add Two Numbers.
初等数学
该问题实际为大数相加问题，若数值过大，暴力求解可能会出现问题。应使用变量carry来跟踪进位，并从包含最低有效位的表头开始模拟逐位相加的过程。
就像在纸上计算两个数字的和那样，我们首先从最低有效位也就是列表l1和l2的表头开始相加。由于每位数字都应当处于0 …… 9的范围内，我们计算两个数字的和时可能会出现 “溢出”。例如，5 + 7 = 12。在这种情况下，我们会将当前位的数值设置为2，并将进位carry=1带入下一次迭代。进位carry必定是0或1，这是因为两个数字相加（考虑到进位）可能出现的最大和为9 + 9 + 1 = 19。[1]
[2]
★ 时间复杂度：\(O(max(m,n))\)
★ 空间复杂的：\(O(max(m,n))+1\)
Elementary Math -python-1
class Solution(object):
def addTwoNumbers(self, l1, l2):
carry=0
dummyHead=ListNode(0)
pointer=dummyHead
while (l1 is not None) or (l2 is not None) or carry==1:
if l1 is not None:
x=l1.val
l1=l1.next
else:
x=0
if l2 is not None:
y=l2.val
l2=l2.next
else:
y=0
sum=x+y+carry
if carry==1:
carry=0
if sum>9:
carry=1
sum%=10
pointer.next=ListNode(sum)
pointer=pointer.next
return dummyHead.next
Runtime: 48 ms, faster than 93.27% of Python online submissions for Add Two Numbers.
Memory Usage: 11.9 MB, less than 47.79% of Python online submissions for Add Two Numbers.
代码可精简部分
1. 用divmod()函数代替判断语句和%=
2.用 ? : 语句代替if else
Elementary Math -python-2
class Solution(object):
def addTwoNumbers(self, l1, l2):
carry=0
dummyHead=ListNode(0)
pointer=dummyHead
while l1 or l2 or carry:
carry,sum=divmod((l1.val if l1 else 0)+(l2.val if l2 else 0)+carry,10)
pointer.next=ListNode(sum)
pointer=pointer.next
l1=l1.next if l1 else None
l2=l2.next if l2 else None
return dummyHead.next
Runtime: 48 ms, faster than 93.27% of Python online submissions for Add Two Numbers.
Memory Usage: 11.8 MB, less than 62.50% of Python online submissions for Add Two Numbers.
代码可更改部分
1. 可用递归代替while循环
Elementary Math + recursion -python
class Solution(object):
def addTwoNumbers(self, l1, l2,dummyHead=ListNode(0),carry=0):
pointer=dummyHead
if l1 or l2 or carry:
carry,sum=divmod((l1.val if l1 else 0)+(l2.val if l2 else 0)+carry,10)
pointer.next=ListNode(sum)
pointer=pointer.next
l1=l1.next if l1 else None
l2=l2.next if l2 else None
self.addTwoNumbers(l1,l2,pointer,carry)
return dummyHead.next
else:
return None
Runtime: 44 ms, faster than 97.92% of Python online submissions for Add Two Numbers.
Memory Usage: 12.1 MB, less than 5.88% of Python online submissions for Add Two Numbers.
♥ 如有错误或有其他解题思路，欢迎指出~~~φ(≧ω≦*)♪
拓展
若链表中的数字不是按逆序存储的呢？[1:1]
示例
输入：(3 -> 4 -> 2) + (4 -> 6 -> 5)
输出：8 -> 0 -> 7
原因：342 + 465 = 807
