Question

我参加了一个竞赛，问题是要从N个项目的数组中找出有效数字的K个数字。

例如

3 -> N (Number of items in Array)

1 2 3 ->A (The array itself constituting N items)

2 -> K (This is the given number of digits to form)

输出应类似于

[0，1] = 12

[0，2] = 13

[1，0] = 21

[1，2] = 23

[2，0] = 31

[2，1] = 32

我猜dynamic problem可以解决此问题的逻辑是什么，但如果能得到帮助，我将感到非常高兴。

Answer 1

我们可以构造一个递归函数，该函数可以从给定的一位整数整数数组中打印出所有可能的K位数字。 Ideone链接：https://ideone.com/RTNz2o

def gen(A, K, arr = []):
    if len(arr) == K:
        print (arr, "=", "".join([str(A[i]) for i in arr]))
    for i in range(0, len(A)):
        val = A[i]
        if val == 0 and len(arr) == 0:
            # We don't want numbers starting with 0
            continue
        if i in arr:
            # We don't want to include the same element again
            continue
        arr.append(i)
        gen(A, K, arr)
        arr.pop()

gen([1, 2, 3], 2)

从给定数组中形成某个数字的子序列

输出应类似于

1 个答案: