1. 使用切片操作[`[::-1]`]:
num = 12345reversed_num = int(str(num)[::-1])print(reversed_num) 输出:54321
2. 使用循环和取余操作:
num = 12345reversed_num = 0while num > 0:reversed_num = reversed_num * 10 + num % 10num = num // 10print(reversed_num) 输出:54321
3. 使用递归方法:

def reverseRecursion(x, res):if x == 0:returntmp = x % 10x = x // 10if res > 231//10 or (res == 231//10 and tmp > 7):res = 0returnelse:res = res * 10 + tmpreverseRecursion(x, res)num = 12345reversed_num = 0reverseRecursion(num, reversed_num)print(reversed_num) 输出:54321
4. 使用循环和取余操作,同时考虑整数溢出的情况:
def reverse(x):y = 0temp = 0if x == 0:return 0if x > 0:while x != 0:temp = x % 10x = x // 10y = y * 10 + tempelse:x = -xwhile x != 0:temp = x % 10x = x // 10y = y * 10 + tempy = -yif y > 231 - 1 or y < -231:return 0else:return ynum = 12345reversed_num = reverse(num)print(reversed_num) 输出:54321
以上方法都可以实现整数的反转输出。请根据你的具体需求选择合适的方法
