Leetcode 算法 - 2. Add Two Numbers
Posted August 16, 2016
问题链接 : 2. Add Two Numbers
问题描述 : You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
使用语言 : Python
需要注意的是: 这里使用的ListNode不是Python内置的list对象, 而是传统(计算机科学)意义上的列表, 通常叫做Linked list (链表), 通常由一系列节点实现, 其每个节点中都持有一个指向下一个节点的引用. 这里就是next属性. 链表分为单向链表和双向链表. 双向链表持有一个指向前一节点的引用.
# 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
"""
x1 , x2 = l1 , l2
results = ListNode ( 0 )
p = results
while True :
if not x1 and not x2 and not p . next :
break
if p . next :
c = getattr ( x1 , 'val' , 0 ) + getattr ( x2 , 'val' , 0 ) + flag . next . val
else :
c = getattr ( x1 , 'val' , 0 ) + getattr ( x2 , 'val' , 0 )
flag = ListNode ( c % 10 )
if c >= 10 :
flag . next = ListNode ( c / 10 )
p . next = flag
p = flag
x1 = x1 . next if isinstance ( x1 , ListNode ) else None
x2 = x2 . next if isinstance ( x2 , ListNode ) else None
return results . next
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