1. 使用切片操作[`[::-1]`]:
num = 12345
reversed_num = int(str(num)[::-1])
print(reversed_num) 输出:54321
2. 使用循环和取余操作:
num = 12345
reversed_num = 0
while num > 0:
reversed_num = reversed_num * 10 + num % 10
num = num // 10
print(reversed_num) 输出:54321
3. 使用递归方法:
def reverseRecursion(x, res):
if x == 0:
return
tmp = x % 10
x = x // 10
if res > 231//10 or (res == 231//10 and tmp > 7):
res = 0
return
else:
res = res * 10 + tmp
reverseRecursion(x, res)
num = 12345
reversed_num = 0
reverseRecursion(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 = 0
temp = 0
if x == 0:
return 0
if x > 0:
while x != 0:
temp = x % 10
x = x // 10
y = y * 10 + temp
else:
x = -x
while x != 0:
temp = x % 10
x = x // 10
y = y * 10 + temp
y = -y
if y > 231 - 1 or y < -231:
return 0
else:
return y
num = 12345
reversed_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