python3 题目 有四个数字:1、2

发布时间:2019-06-08 21:13:40编辑:auto阅读(1917)

    方法一:for循环遍历

    counter=0
    for i in range(1,5):
        for j in range(1,5):
            for k in range(1,5):
                if i !=j and j !=k and k !=i:
                    print("{}{}{}".format(i,j,k),end=" ")
                    counter +=1
    print("")
    print("共{}种组合".format(counter))

    方法二:itertools中的permutations即可

    counter=0
    from itertools import permutations
    for i in permutations([1,2,3,4],3):
        print("{}{}{}".format(i[0],i[1],i[2]),end=" ")
        counter +=1
    print("")
    print("共{}种组合".format(counter))

    效果:

     

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