计算范围 (0,n] 中数字“x”的出现次数

Question

所以我正在尝试编写一个 python function，它接受两个 arguments，n 和 num，并计算 0 和 num 之间“n”的出现次数。 例如，

countOccurrences(15,5)应该是2 。

countOccurrences(100,5)应该是20 。

我对这个问题做了一个简单的迭代解决方案：

def countOccurrences(num,n):
  count=0
  for x in range(0,num+1):
    count += countHelper(str(x),n)
  return count

def countHelper(number,n):
  count=0
  for digit in number:
    if digit==n:
      count += 1
  return count

如果我尝试调用countOccurrences(100000000000,5) ，这会遇到明显的问题。 我的问题是如何提高效率？ 我希望能够“相当”快速地处理问题，并避免出现 memory 个错误。 这是我尝试执行此操作的递归解决方案的第一次传递：

def countOccurence(num, n):
  if num[0]==n:
    return 1
  else:
    if len(num) > 1:
      return countOccurence(num[1:],n) + countOccurence(str((int(num)-1)),n)
    else:
      return 0

Answer 1

除非max_num足够小以适合C long ， max_num这不会遇到任何内存问题。 基本上，它仍然是蛮力算法，尽管已针对Python进行了优化。

def count_digit(max_num, digit):
    str_digit = str(digit)
    return sum(str(num).count(str_digit) for num in xrange(max_num+1))

Answer 2

我已经解决了我的解决方案，希望它符合您的规格。 让我们看一下第一个辅助函数：

def splitNumber(num):
    temp = str(number)
    nums = list(temp)
    return nums

此函数创建一个字符串列表，列出所有数字输入中的各个数字。 例如，

splitNumber(100)

会返回：

['1', '0', '0']

在这里，您可以调用main函数并使用此main函数测试每个数字：

def countOccurences(num, n):
    count = 0
    for x in range(0, (num + 1)):
        temp = splitNumber(x)
        for x in range(len(temp)):
            if (temp[x] == str(n)):
                count = count + 1
    return count

哪个应该给出期望的输出。 让我知道这是否适合您！

Answer 3

请参阅： https://math.stackexchange.com/a/1229292/974150

在 python 中：

def counts_of_i_bf(n, i):
    """Counts the number of occurences in a range [0 .. n] of
        the digit i [0...9]

    Args:
        n ([int]): upper value of range [0 ... n]
        i ([type]): digit looking for [0.. 9]

    Returns:
        [int]: occurences of i in the range [0...n]
    """
    return sum(str(d).count(str(i)) for d in range(n + 1))

def counts_of_i_dp(n, i):
    """Counts the number of occurences in a range [1 .. n] of
        the digit i [1...9] by implementing the recurrence
        relation:
                        | ak.10^(k-1) + fi(b)           if a < i
        fi(a.10^k +b) = | ak.10^(k-1) + 1 + fi(b) + b   if a == i
                        | (ak + 10).10^(k-1) + fi(b)    if a > i

    see: https://math.stackexchange.com/a/1229292/974150
    Args:
        n ([int]): upper value of range [1 ... n]
        i ([type]): digit looking for [1.. 9]

    Returns:
        [int]: occurences of i in the range [0...n]
    """
    som = 0
    while n > 0:
        k = int(log10(n))
        a = n // 10**k
        b = n - a * 10**k
        if a < i:
            som += a*k*10**(k-1)
        elif a == i:
            som += a*k*10**(k-1) + 1 + b
        else:
            som += (a*k + 10)*10**(k-1)
        n = b
        
    return int(som)

def counts_of_0(n):
    """Counts the number of occurences in a range [1 .. n] of
        the digit0 by implementing:
        Tn = (k + 1)*(b + 1 + (a - 1)10^k) + ∑ 9*s*10(s - 1) for s=1.. k\
        f0(n) = Tn -∑ 9s.10^(s-1) for s=1..9
  
        see: https://math.stackexchange.com/a/1229292/974150
    Args:
        n ([int]): upper value of range [1 ... n]

    Returns:
        [int]: occurences of 0 in the range [1...n]
    """
    k = int(log10(n))
    a = n // 10**k
    b = n - a * 10**k
    Tn = (k + 1)*(b + 1 + (a - 1)*10**k) + sum(9*s*10**(s - 1) for s in range(1, k + 1))
    return Tn - sum(counts_of_i_dp(n, i) for i in range(1, 10)) + 1 # was one of


def counts_of_i(n, i):
    """Counts the number of occurences in a range [0 .. n] of
        the digit i [0...9]
        

    Args:
        n ([int]): upper value of range [0 ... n]
        i ([type]): digit looking for [0.. 9]

    Returns:
        [int]: occurences of i in the range [0...n]
    """
    if n == 0: return 1 if i == 0 else 0
    if i == 0: return counts_of_0(n)
    return counts_of_i_dp(n, i)

assert all(counts_of_i_bf(i, d) == counts_of_i(i, d) for i in range(1_001) for d in range(10))