problem description:

you should change the given digits string into possible letter string according to the phone keyboards.

i.e.

input '23'

output ['ad','ae','af','bd','be','bf','cd','ce','cf']

the python solution

first you should realize this a iteration process, so you can use many iterate process to handle this problem.reduce fuction is a important iteration function in python.reduce(function , iterator, start),this is it's base form,the first function must have two parameter, the first parameter will be used to record the result,and the other just to fetch the number in the iterator.start can be omitted, then the fisrt result fetch from the iterator the first num,if not the start means the original result.After know the reduce fuction, we know can use it to solve this problem.

in python:

lists = ['a','b']

for in in 'abc':

  lists += i

return lists

then the lists will be ['aa','ab','ac','ba','bb','bc'],base on this feacture, so give the below solution

class Solution(object):
def letterCombinations(self, digits):
"""
:type digits: str
:rtype: List[str]
"""
if digits == '': return []
phone = {'':'abc','':'def','':'ghi','':'jkl',
'':'mno','':'pqrs','':'tuv','':'wxyz'}
return reduce(lambda x,y:[a+b for a in x for b in phone[y]],digits, [''])

this solution is so smart.Thanks to huxley  publish such a great  method to deal with this issue.

I f you still can not understand this method, there is another simple way.

class Solution(object):
def letterCombinations(self, digits):
"""
:type digits: str
:rtype: List[str]
"""
if len(digits) == 0:
return []
phone = {'':'abc','':'def','':'ghi','':'jkl',
'':'mno','':'pqrs','':'tuv','':'wxyz'} result = ['']
for i in digits:
temp = []
for j in result:
for k in phone[i]:
temp.append(j + k)
result = temp
return result

phone number的更多相关文章

  1. JavaScript Math和Number对象

    目录 1. Math 对象:数学对象,提供对数据的数学计算.如:获取绝对值.向上取整等.无构造函数,无法被初始化,只提供静态属性和方法. 2. Number 对象 :Js中提供数字的对象.包含整数.浮 ...

  2. Harmonic Number(调和级数+欧拉常数)

    题意:求f(n)=1/1+1/2+1/3+1/4-1/n   (1 ≤ n ≤ 108).,精确到10-8    (原题在文末) 知识点:      调和级数(即f(n))至今没有一个完全正确的公式, ...

  3. Java 特定规则排序-LeetCode 179 Largest Number

    Given a list of non negative integers, arrange them such that they form the largest number. For exam ...

  4. Eclipse "Unable to install breakpoint due to missing line number attributes..."

    Eclipse 无法找到 该 断点,原因是编译时,字节码改变了,导致eclipse无法读取对应的行了 1.ANT编译的class Eclipse不认,因为eclipse也会编译class.怎么让它们统 ...

  5. 移除HTML5 input在type="number"时的上下小箭头

    /*移除HTML5 input在type="number"时的上下小箭头*/ input::-webkit-outer-spin-button, input::-webkit-in ...

  6. iOS---The maximum number of apps for free development profiles has been reached.

    真机调试免费App ID出现的问题The maximum number of apps for free development profiles has been reached.免费应用程序调试最 ...

  7. 有理数的稠密性(The rational points are dense on the number axis.)

    每一个实数都能用有理数去逼近到任意精确的程度,这就是有理数的稠密性.The rational points are dense on the number axis.

  8. [LeetCode] Minimum Number of Arrows to Burst Balloons 最少数量的箭引爆气球

    There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided ...

  9. [LeetCode] Number of Boomerangs 回旋镖的数量

    Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of po ...

  10. [LeetCode] Number of Segments in a String 字符串中的分段数量

    Count the number of segments in a string, where a segment is defined to be a contiguous sequence of ...

随机推荐

  1. 推荐一本不错的书《Sencha Ext JS 5 Bootcamp in a Book》

    原文:https://www.createspace.com/5425618 看了一下该书目录,感觉不错,Ext JS 5的重点内容都提及了,确实是一本学习Ext JS 5的好书,唯一遗憾的地方就是太 ...

  2. Java虚拟机结构

    一.JVM主要的结构如下: 二.各个区域功能介绍 1).方法区(Method Area):         (1)用于存储虚拟机加载的类信息.常量.静态变量等,是各个线程共享的内存区域:       ...

  3. Dynamics CRM OData方式进行增删改查时报错的问题

    今天在通过OData终结点update记录的时候报"Error processing request stream. The request should be a valid top-le ...

  4. tomcat集群实现源码级别剖析

    随着互联网快速发展,各种各样供外部访问的系统越来越多且访问量越来越大,以前Web容器可以包揽接收-逻辑处理-响应整个请求生命周期的工作,现在为了构建让更多用户访问更强大的系统,人们通过不断地业务解耦. ...

  5. spring 注解模式 详解

    Spring基于注解实现Bean定义支持如下三种注解: Spring自带的@Component注解及扩展@Repository.@Service.@Controller,如图12-1所示: JSR-2 ...

  6. hibernate关联对象的增删改查------增

    本文可作为,北京尚学堂马士兵hibernate课程的学习笔记. 这一节,我们看看hibernate关联关系的增删改查 就关联关系而已,咱们在上一节已经提了很多了,一对多,多对一,单向,双向... 其实 ...

  7. 漫谈程序员(十八)windows中的命令subst

    漫谈程序员(十八)windows中的命令subst 用法格式 一.subst [盘符] [路径]  将指定的路径替代盘符,该路径将作为驱动器使用 二.subst /d 解除替代 三.不加任何参数键入  ...

  8. 类成员函数后边加const

    本文主要整理自stackoverflow上的一个对问题Meaning of “const” last in a C++ method declaration?的回答. 测试1 对于下边的程序,关键字c ...

  9. 【一天一道LeetCode】#5 Longest Palindromic Substring

    一天一道LeetCode系列 (一)题目 Given a string S, find the longest palindromic substring in S. You may assume t ...

  10. ITU-T Technical Paper: QoS 的参数(非常的全,共计88个)

    本文翻译自ITU-T的Technical Paper:<How to increase QoS/QoE of IP-based platform(s) to regionally agreed ...