面试的一道编程题

foxiangzun

def numCount(num1) :
        list1 = list(str(num1))
        dic = {}
        set1 = set(list1)
        for i in set1 :
                dic[i] = 0
        for i in list1 :
                for j in dic :
                        if i == j :
                                dic[i] += 1
        print(dic)

永恒的蓝色梦想

不明白，按自己理解的来了

def func(lis):return{key:lis.count(key) for key in set(lis)}

Jung

def Numcount(number):
    String = str(number)
    L = []
    for num in String:
        if (String.count(num)>1) and (num not in L):
            print("The duplicated number is %s .And times %d" %(num,String.count(num)))
            L.append(num)

nononoyes

str = "123451112"
newstr = ""
for i in range(len(str)):
    if str.count(str[i])>1 and newstr.count(str[i])<1:
        newstr+=str[i]
        print("%s 出现的次数%s"%(str[i],str.count(str[i])))

其实可以用字典的KEY是唯一的来存储数字，用value来存储该数字重复的次数

今天的我更强了

num=[int(i) for  i in input('请输入一段数字并用‘,’隔开：').split(',')]
list1=[]
for each in num:
    if each not in list1:
        list1.append(each)
    else:
        print(each,'重复了：%d次'%(num.count(each)))