记录面试过程中的2个算法知识

问题一

Prime number (质 数)

  Given a positive integer number n. Determine whether n is a prime number or not. (You may implement your program in any programming language.)
  (给定一个正整数n。判断n是否是素数(你可以用任何编程语言来实现你的程序)

  我当时给出的解答: 发现如果使用 111111111111529  这个大数;直接就卡死了;

def is_prime(number):
  count = 0
  if number<= 1:
    print('不是')
  else:
      for i in range(1, number+1):
        if number%i==0:
          count+=1
  if count > 2:
      print(False)
  else:
      print(True)
if __name__ == "__main__":
  # is_prime(6)
  is_prime(111111111111529)

 

于是我给优化了一下;

def isPrime(n):
    start_time = time.time()
    import math
    if n <= 1:
        return False
    for i in range(2,int(math.sqrt(n))+1,6):
        if n % i == 0 or n % i+2 ==0:return '不是质数'
    consume_time = time.time() - start_time
    return ['是质数',consume_time]

if __name__ == "__main__":
  print(isPrime(2))

上面两种方法的时间复杂度都是为 O(n); 但是面试官和我说 可以改成 O(1);我也是一脸懵逼;难道是用递归来弄么;如果有人可以用O(1)的时间复杂度来解决 请评论一下;让我涨涨见识;

问题二

2.Largest number

# // 2.1 Given a list of non negative integers, arrange them such that they form the largest number. (You may implement your program in pseudocode or any programming language.)
  (给定一个非负整数列表,将它们排列成最大的数(您可以用伪代码或任何编程语言实现您的程序。)

def sum_number1(numList):
    '''
    ****给定的是一个list
    ****做一个非整数判断
    '''
    auxiliary = [str(i) for i in numList if str(i).isdigit()]
    if not auxiliary:
        return ''
    length = len(auxiliary)
    for index1 in range(length):
        for index2 in range(length - index1 - 1):
            if auxiliary[index2] + auxiliary[index2 + 1] < auxiliary[index2 + 1] + auxiliary[index2]:
                auxiliary[index2], auxiliary[index2 + 1] = auxiliary[index2 + 1], auxiliary[index2]
    result = ''
    for index3 in range(length):
        result = result + auxiliary[index3]
    print(result)
if __name__ == "__main__":
    # sum_number([1,29,'4123',59,'a'])
    sum_number1([19, 49, '3123', 59, 'a'])

 

 

第二题的时间复杂度O(n^2);  

 

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