CheckIO 题目解法(1)

CheckIO是一个用Python语言解决问题的网站，有一些好玩的问题可以试试解决一下.看起来很像是一场冒险，赶快开始吧！

1.  Extra dashes 去除多余的"-"比如输入"I---Like---Python"输出"I-Like-Python" 可以这么写:

def checkio(line):
    return '-'.join([x for x in line.split('-') if x != ''])

2. Non-unique Elements 去除一个list中只出现一次的数字 可以这么写:

def checkio(data): 
    return [i for i in data if data.count(i) != 1]

3. Median 找出一个list的中位数字,奇数个数字就是排序后的中间数，偶数个数字就是排序后的中间两个数字取均值 可以这么写:

def checkio(data):
    length = len(data)
    data.sort()
    if length % 2 == 0:
        res = (data[int(length / 2) - 1] + data[int(length / 2)]) / 2.0
    else:
        res = data[int(length / 2)]
    return res

4.House password  判断口令是不是够strong,长度不少于10，至少一个数字，一个大写字母，一个小写字母 可以这么写:

import string

def checkio(data):
    condtions = [False] * 4
    if len(data) >= 10:
        condtions[0] = True
    for i in data:
        if i in string.digits:
            condtions[1] = True
        elif i in string.ascii_lowercase:
            condtions[2] = True
        elif i in string.ascii_uppercase:
            condtions[3] = True
    return all(condtions)

 5.The Most Wanted Letter  找出来最想要的字母: 出现次数最多，如果相同按照字母顺序a-z 优先级 可以这样做:

import string

def checkio(text):
    low_text = text.lower()
    # print low_text
    index = 0
    for c in range(len(low_text)):
        if low_text[c] in string.ascii_lowercase:
            if low_text.count(low_text[c]) > low_text.count(low_text[index]):
                index = c
            elif low_text.count(low_text[c]) == low_text.count(low_text[index]):
                if string.ascii_lowercase.index(low_text[c]) < string.ascii_lowercase.index(low_text[index]):
                    index = c
    return low_text[index]

 6.Speech Module  从0-999的数字翻译成英文 最简单的就是挨个写一遍 不过太麻烦 可以这样做:

def checkio(num):
        res = ''
        one_num_dict = {'0': 'zero', '1': 'one', '2': 'two', '3': 'three', '4': 'four','5': 'five', '6': 'six', '7': 'seven', '8': 'eight', '9': 'nine',
                        '10': 'ten', '11': 'eleven', '12': 'twelve', '13': 'thirteen',
                        '14': 'fourteen','15': 'fifteen', '16': 'sixteen',
                        '17': 'seventeen', '18': 'eighteen', '19': 'nineteen',
                        '20': 'twenty', '30': 'thirty', '40': 'forty', '50': 'fifty',
                        '60': 'sixty' , '70': 'seventy', '80': 'eighty', '90': 'ninety'}
        # keys = one_num_dict.keys()
        str_num = str(num)
        if len(str_num) == 1:
                res = one_num_dict[str_num]
        elif len(str_num) == 2:
                if str_num == '00':
                        return ''
                if str_num[0] == '1':
                        res =  one_num_dict[str_num]
                else:
                        if str_num[1] == '0':
                                res = one_num_dict[str_num]
                        else:
                                res  = one_num_dict[str_num[0] + '0'] + ' ' + checkio(int(str_num[1]))
        else:
                res = one_num_dict[str_num[0]] + ' ' + 'hundred ' + checkio(int(str_num[1:3]))
                if res.endswith('zero'):
                        res = res[0:-5].strip()
        return res.strip()

 7.Find-Ip 从一个字串中匹配出来IP地址,不含前导0,最简单的做法是用正则表达式 不过自己实现的话 可以这么写:

def checkio(ip_string):
        res = []
        words = ip_string.split(' ')
        words = filter(lambda x: len(x.split('.')) == 4, words)
        for word in words:
                is_ip = True
                nums = word.split('.')
                if not nums[0].isdigit() or int(nums[0]) == 0:
                        is_ip = False
                for num in nums:
                        if not num.isdigit():
                                is_ip = False
                                break
                        if (len(num) > 3) or (not (int(num) >= 0 and int(num) <= 255)):
                                is_ip = False
                                break
                if is_ip:
                        res.append(word)
        return res

 8.Brackets 监测一个表达式的括号是否正确的匹配了，可以使用一个list当栈来检查：

def checkio(expr):
    a_stack = []
    for i in expr:
        if i in '({[':
            a_stack.append(i)
        elif i in '}])':
            if a_stack == []:
                return False
            else:
                poped = a_stack.pop()
                if poped == '(' and i != ')':
                    return False
                elif poped == '[' and i != ']':
                    return False
                elif poped == '{' and i != '}':
                    return False
    return len(a_stack) == 0

 9. Morse-Clock 转换10进制到2进制，比较简单 不过我写的效率...:

def checkio(timestr):
    numlist = timestr.split(':')
    alist = []
    numlist = ["{:0>2d}".format(int(i)) for i in numlist]
    for i in numlist:
        for c in i:
            alist.append(bin(int(c)))
    res = ''
    res += "{:0>2d}".format(int(alist[0].replace('0b',''))) + ' '
    res += "{:0>4d}".format(int(alist[1].replace('0b', ''))) + ' : '
    res += "{:0>3d}".format(int(alist[2].replace('0b', ''))) + ' '
    res += "{:0>4d}".format(int(alist[3].replace('0b', ''))) + ' : '
    res += "{:0>3d}".format(int(alist[4].replace('0b', ''))) + ' '
    res += "{:0>4d}".format(int(alist[5].replace('0b', '')))
    res = res.replace('1', '-')
    res = res.replace('0', '.')
    return res

 10.Striped Words 如果有两个字母都是元音和辅音的话不算，其他算一个：

import string
import re

def checkio(text):
    re_str = "[?!;,. ]"
    vowels = 'AEIOUY'
    con = 'BCDFGHJKLMNPQRSTVWXZ'
    count = 0
    res = []
    words = [i for i in re.split(re_str, text) if i and len(i) != 1]
    for word in words:
        length = len(word) - 1
        ismyword = True
        for i in xrange(length):
            if (word[i].upper() in vowels and word[i+1].upper() in vowels) 
               or (word[i].upper() in con and word[i+1].upper() in con) 
               or (word[i] in string.digits):
                ismyword = False
                break
        if ismyword:
            count += 1
    return count