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  • Python练习(一)

    分别用for、while、递归求列表的和

    >>> def list_sum(argument_list):
        sum = 0
        for arg in argument_list:
            sum += arg
        print('for =>',sum)
    
        sum = 0
        count = len(argument_list)
        while count > 0:
            sum += argument_list[count-1]
            count -= 1
        print('while =>',sum)
        
        def recursion_sum(arg_list):
            if len(arg_list) == 1:
                return     arg_list[0]
            else:
                return     arg_list[len(arg_list)-1] + recursion_sum(arg_list[:len(arg_list)-1])
        sum = recursion_sum(argument_list)
        print('recursion =>', sum)
    
        
    >>> list_sum([1, 2, 3, 4, 5])
    for => 15
    while => 15
    recursion => 15

    交错合并列表函数

    >>> def interlaced_combination_fun(a_list, b_list):
        result = [];
        for i in range(min(len(a_list), len(b_list))):
            if i < len(a_list):
                result.append(a_list[i])
            if i < len(b_list):
                result.append(b_list[i])
        print(result)
    
        
    >>> interlaced_combination_fun([1, 3, 5, 7, 9], [2, 4, 6, 8])
    [1, 2, 3, 4, 5, 6, 7, 8]

     给定非负整数列表中数字排列成最大数字

    >>> def from_list_get_maxnum(arg_list):
        for i in range(len(arg_list)):
            for j in range(i):
                for k in range(min(arg_list[i]//10+1, arg_list[j]//10+1)):
                    if int(str(arg_list[j])[k]) < int(str(arg_list[i])[k]):
                        arg_list[j], arg_list[i] =  arg_list[i],  arg_list[j]
                        break
        result = ''
        for arg in arg_list:
            result = result + str(arg)
        result = int(result)
        return result
    
    >>> from_list_get_maxnum([50, 2, 1, 9])
    95021

    计算前100位的斐波那契函数

    >>> def fibonacci(x):
        a, b = 0, 1
        print(a, end=' ')
        while 1:
            if a+b > 100:
                break
            a, b = b, a+b
            print(b, end = ' ')
    
            
    >>> fibonacci(100)
    0 1 2 3 5 8 13 21 34 55 89 

    123456789顺序不能变,数字之间插入加减或什么都不插入,使结果总在100,输出所有可能性

    >>> def fun(numbers, build):
        if(numbers):
            for char in ["+","","-"]:
                tmp = build + char + numbers[0]
                fun(numbers[1:], tmp)
        elif(eval(build) == 100):
            print(build)
    
            
    >>> fun("23456789","1")
    1+2+34-5+67-8+9
    1+2+3-4+5+6+78+9
    1+23-4+5+6+78-9
    1+23-4+56+7+8+9
    12+3+4+5-6-7+89
    12+3-4+5+67+8+9
    123+45-67+8-9
    123+4-5+67-89
    123-45-67+89
    123-4-5-6-7+8-9
    12-3-4+5-6+7+89
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  • 原文地址:https://www.cnblogs.com/liyonghua/p/7772897.html
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