两数相加
2018-10-02 本文已影响0人
3ni
给定两个非空链表来表示两个非负整数。位数按照逆序方式存储,它们的每个节点只存储单个数字。将两数相加返回一个新的链表。
你可以假设除了数字 0 之外,这两个数字都不会以零开头。
示例:
输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出:7 -> 0 -> 8
原因:342 + 465 = 807
# 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
"""
re = None
a = l1
b = l2
c = re
yu = 0
jin = 0
while a or b:
t = ListNode(0)
if re is None:
re = t
c = re
else:
c.next = t
c = c.next
if a and b:
yu = (a.val + b.val + jin) % 10
jin = (a.val + b.val + jin) / 10
c.val = yu
elif a:
yu = (a.val + jin) % 10
jin = (a.val + jin) / 10
c.val = yu
elif b:
yu = (b.val + jin) % 10
jin = (b.val + jin) / 10
c.val = yu
if a:
a = a.next
if b:
b = b.next
if jin:
t = ListNode(0)
c.next = t
c = c.next
c.val = jin
return re
思路:每位上分别相加(比如个位就个位相加,十位就十位相加),注意还要加上低位的进位,所以每位都要记录是否有进位,然后把新生成的位加入一个新的链表,重复这个过程。
注意:要考虑两个链表长度不等,最高位的进位比如(999 + 100 = 1099变成四位数),第一个链表节点的处理方式等