查找给定范围内给定数字的整数个数的算法

如果我以列表的形式得到完整的数字集list我想知道它们在给定范围内可以形成多少个（有效）整数[A, B]，我可以使用什么算法来有效地完成它？

例如，给定一个数字列表（包含重复项和零）list={5, 3, 3, 2, 0, 0},我想知道这个范围内可以组成多少个整数[A, B]=[20, 400]包括的。例如，在这种情况下，20, 23, 25, 30, 32, 33, 35, 50, 52, 53, 200, 203, 205, 230, 233, 235, 250, 253, 300, 302, 303, 305, 320, 323, 325, 330, 332, 335, 350, 352, 353都是有效的。

Step 1: Find the number of digits your answers are likely to fall in. In your 
        example it is 2 or 3.

Step 2: For a given number size (number of digits)

    Step 2a: Pick the possibilities for the first (most significant digit). 
    Find the min and max number starting with that digit (ascend or descending
    order of rest of the digits). If both of them fall into the range:
        step 2ai: Count the number of digits starting with that first digit and
        update that count
    Step 2b: Else if both max and min are out of range, ignore. 
    Step 2c: Otherwise, add each possible digit as second most significant digit
    and repeat the same step 

根据您的案例解决：

对于数字大小为 2 即 __：

0_ : Ignore since it starts with 0
2_ : Minimum=20, Max=25. Both are in range. So update count by 3 (second digit might be 0,3,5)
3_ : Minimum=30, Max=35. Both are in range. So update count by 4 (second digit might be 0,2,3,5)
5_ : Minimum=50, Max=53. Both are in range. So update count by 3 (second digit might be 0,2,3)

对于尺寸 3：

0__ : Ignore since it starts with 0
2__ : Minimum=200, max=253. Both are in range. Find the number of ways you can choose 2 numbers from a set of {0,0,3,3,5}, and update the count.
3__ : Minimum=300, max=353. Both are in range. Find the number of ways you can choose 2 numbers from a set of {0,0,2,3,5}, and update the count.
5__ : Minimum=500, max=532. Both are out of range. Ignore.

一个更有趣的情况是最大限制为 522（而不是 400）：

5__ : Minimum=500, max=532. Max out of range.
    50_: Minimum=500, Max=503. Both in range. Add number of ways you can choose one digit from {0,2,3,5}
    52_: Minimum=520, Max=523. Max out of range.
        520: In range. Add 1 to count.
        522: In range. Add 1 to count.
        523: Out of range. Ignore.
    53_: Minimum=530, Max=532. Both are out of range. Ignore.



def countComb(currentVal, digSize, maxVal, minVal, remSet):
    minPosVal, maxPosVal = calculateMinMax( currentVal, digSize, remSet)
    if maxVal>= minPosVal >= minVal and maxVal>= maxPosVal >= minVal
        return numberPermutations(remSet,digSize, currentVal)
    elif minPosVal< minVal and maxPosVal < minVal or minPosVal> maxVal and maxPosVal > maxVal:
        return 0
    else:
        count=0
        for k in unique(remSet):
            tmpRemSet = [i for i in remSet]
            tmpRemSet.remove(k)
            count+= countComb(currentVal+k, digSize, maxVal, minVal, tmpRemSet)
        return count

在你的情况下：countComb('',2,400,20,['0','0','2','3','3','5']) + countComb('',3,400,20,['0','0','2','3','3','5'])会给出答案。

def calculateMinMax( currentVal, digSize, remSet):
    numRemain = digSize - len(currentVal)
    minPosVal = int( sorted(remSet)[:numRemain] )
    maxPosVal = int( sorted(remSet,reverse=True)[:numRemain] )
    return minPosVal,maxPosVal

numberPermutations(remSet,digSize, currentVal): Basically number of ways 
you can choose (digSize-len(currentVal)) values from remSet. See permutations
with repeats.