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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
我的写法是:
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
carry = 0 #表示是否有进位
start = l1 #指针起始位置
pre = None #计算的前一个节点
while (l1 != None) and (l2 != None):
sum = l1.val + l2.val + carry #计算位数的和
#判断该位数计算和之后是否需要向下一位进位
if sum >= 10:
l1.val = sum - 10
carry = 1
else:
l1.val = sum
carry = 0
pre = l1
l1 = l1.next
l2 = l2.next
#当两个链表长度不一致时的处理
#1、当链表2比链表1长时
if(l2 != None):
pre.next = l2
#将l1指向l2剩下的节点进行处理
l1 = pre.next
#2、当链表1比链表2长时的处理
#此时链表2的指针指向末尾,所以只需要判断是否存在进位即可得出两数之和
while l1 != None:
if carry == 1:
if l1.val==9:
l1.val = 0
carry = 1
else:
l1.val += 1
carry = 0
pre = l1
l1 = l1.next
#此时两个链表已经计算完毕,只需要判断最后的计算结果是否需要添加新节点存储
if carry == 1:
pre.next = ListNode(1)
return start
大神的写法是:
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
carry = 0
root = n = ListNode(0)
while l1 or l2 or carry:
v1 = v2 = 0
if l1:
v1 = l1.val
l1 = l1.next
if l2:
v2 = l2.val
l2 = l2.next
carry, val = divmod(v1+v2+carry, 10)
n.next = ListNode(val)
n = n.next
return root.next