Python3个位十位对调 Python3实现个位数字和十位数字对调, 其乘积不变

count = 0

for i in range(11, 100):
  first_left = i // 10
  first_right = i % 10

  if first_left == first_right:  #减少程序循环次数，因为后面的 for j in range(i + 1, 100)
    continue    # 已经解决了11、22 重复出现的不在此列 和 13*62与62*13不在队列

  for j in range(i + 1, 100):   #遍历出数字并找出个位和十位
    second_left = j // 10
    second_right = j % 10

    if second_left == second_right:  #同理减少循环次数
      continue

    newfirst = first_right * 10 + first_left
    newsecond = second_right * 10 + second_left

    if i * j == newfirst * newsecond: #进行if判断
      print('{} * {} = {} * {}'.format(i, j, newfirst, newsecond))

      count += 1

print(count)

class Test:
  def reverseInteger(self, number):
    g = number%10     #取出个位数
    s = (int(number/10))%10  #取出十位数
    b = int(number/100) #取出百位数
    return g*100+s*10+b
if __name__ == "__main__":
  ts = Test()
  print (ts.reverseInteger(123)) #输出结果：321

class Test:
  def reverseInteger(self, number):
    number = str(number) # 模拟入栈
    l = list(number)
    result = ""
    while len(l) > 0:
      result += l.pop() # 模拟出栈
    return int(result)
if __name__ == "__main__":
  ts = Test()
  print (ts.reverseInteger(123)) # 输出结果：321

class Test:
  def reverseInteger(self, number):
    number=str(number)
    result=number[::-1] #python中切片的特殊用法
    result=(int(result))
    return result
if __name__ == "__main__":
  ts = Test()
  print (ts.reverseInteger(123))

class Test:
  def reverseInteger(self, x):
    plus_minus = ""
    reverse_x = ""
    if x < 0:
      plus_minus = "-"
      x = -x
    for i in str(x):
      reverse_x = i + reverse_x
    reverse_x = plus_minus + reverse_x
    if int(reverse_x) > pow(2, 31) - 1 or int(reverse_x) < pow(-2, 31):
      return 0
    return int(reverse_x)

if __name__ == "__main__":
  ts = Test()
  print (ts.reverseInteger(123)) #输出结果：321