暴力(python)

输出由1,2,3,4组成的互不相同且无重复的三位数!

#方式一

lst = ['1', '2', '3', '4']

res = []

for i in lst:

    for j in lst:

        for k in lst:

            # print(set(i + j + k))

            if len(set(i + j + k)) == 3:

                print(set(i + j + k))

                res.append(int(i + j + k))

print(res)

print(len(res))

#方式二

count = 0

for i in range(1, 5):

    for j in range(1, 5):

        for k in range(1, 5):

            if (i != j) and (i != k) and (j != k):

                count += 1

print(count)

#方式三

lst = "1234”

res = 0

for i in lst:

    for j in lst:

        for k in lst:

            # print(set(i + j + k))

            if len(set(i + j + k)) == 3:

                print(set(i + j + k))

                res += 1

print(res)

#a 与 b必须是两个相同类型的可迭代对象

a = "1"

b = "2"

print(set(a + b))  # {'1', '2'}

a = "123"

b = "2"

print(set(a + b))  # {'1', '2', '3'}

a = ['1', '3', 2]

b = ['2']

print(set(a + b))  # {'2', '3', 2, '1'}

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转载自www.cnblogs.com/xiao-xue-di/p/9644859.html