从给定的输入中打印缺失的数字

Question

按升序排列的数字作为输入作为字符串给出。程序应该打印丢失的数字？

EX:

INPUT

567568600601602

OUTPUT

569

EXPLANATION

567,568,600,601,602 are the numbers in sequence and 569 is the missing number

INPUT

1112131516

OUTPUT

14

EXPLANATION

11,12,13,15,16 are the numbers in sequence and 569 is the missing

我的代码

a=input()
a=a.rstrip()
b=list(set(a))
l=[]
p=[]
c=0
for i in range(len(b)):
    if a.count(b[i])>1:
        c+=1
j=0
while j<=len(a):
    l.append(a[j:j+c+1])
    j=j+c+1
del l[-1]
k=int(l[0])
while k<=int(l[-1]):
    p.append(k)
    k=k+1
for u in range(len(p)):
    if str(p[u]) not in l:
        print(p[u])
        break

我的查询：

如果以字符串形式输入 2 位或 4 位数字，我的程序无法找到丢失的数字。

input 6768707172
output of my program 677
expected output 69

如何检查字符串是否有 2 位或 3 位或 4 位数字并相应地分解它们？

Answer 1

使用伪代码评估算法。识别重要概念并创建带有参数的函数。给出有意义的名字。用铅笔和纸运行迭代。如果可行，请改进算法。确定支持您的用例的数据结构。

当前的代码不是为人眼编写的，但您是在向人类寻求帮助。

Answer 2

我只是为了好玩而采用这种方法，它似乎有效（我没有控制错误的输入异常，它肯定可以改进）。我基本上从 1 个字符长度开始，如果我是对的，请尝试。

希望对你有帮助：

MAX_LEN = 6

def findMissingNumber(s):
    print 'Input received: ' + s

    for digitsNumber in range(1,MAX_LEN+1):
        print 
        # Initializing stuff
        index = digitsNumber
        currentNumber = int(s[:digitsNumber])
        exit = False
        missing = None

        while not exit and index < len(s):
            # Store expected next number
            nextNumber = currentNumber + 1
            # Store expected next number length
            nextNumberLen  = len(str(nextNumber))
            # Store next number in the provided string
            # based the lenght of the expected number
            nextStringNumber = int(s[index : index + nextNumberLen])
            print 'Have ' + str(currentNumber) + ', expecting ' + str([nextNumber,nextNumber+1]) + ' and got ' + str(nextStringNumber)
            # Check if number gotten is the next or the following
            if nextStringNumber in [nextNumber, nextNumber+1]: 
                # Check if number is not the next
                # (means there is a number missing)
                if nextStringNumber != nextNumber:
                    # Check if there was a previous missing number
                    if not missing:
                        print 'Found missing ' + str(nextNumber)
                        missing = nextNumber
                    # If there was, exit  and forget missing number
                    # (only 1 missing number allowed)
                    else:
                        print 'More than 1 missing, exit'
                        missing = None
                        exit = True
                # Set stuff for next iteration
                currentNumber = nextStringNumber
                index += nextNumberLen
            # If end of string, exit
            else:
                print 'End of string, exit'
                exit = True
        # If only 1 missing found, return it
        if missing:
            print 'Returning ' + str(missing)
            return missing
        print 'Going to next number lenght'
    # If no missing found, return -1
    return -1

findMissingNumber('6768707172')

因此而给予：

Input received: 6768707172

Have 6, expecting [7, 8] and got 7
Have 7, expecting [8, 9] and got 6
End of string, exit
Going to next number lenght

Have 67, expecting [68, 69] and got 68
Have 68, expecting [69, 70] and got 70
Found missing 69
Have 70, expecting [71, 72] and got 71
Have 71, expecting [72, 73] and got 72
Returning 69