当前位置:网站首页>[LeetCode] 整数反转【7】
[LeetCode] 整数反转【7】
2022-07-05 04:55:00 【山茶花开时。】
问题: 给你一个32位的有符号整数x,返回将x中的数字部分反转后的结果
如果反转后整数超过32位的有符号整数的范围[-(2**31), 2**31-1] ,就返回0
示例1
输入: x = 123
输出: 321
示例2
输入: x = -123
输出: -321
示例3
输入: x = 120
输出: 21
示例4
输入: x = 0
输出: 0
# 解法1
def reverse(x):
if x == 0:
return 0
if x > 0:
str_x = str(x)
str_x = str_x[::-1]
x = int(str_x)
if x > 2**31 - 1:
return 0
if x < 0:
str_x = str(x)
str_x = str_x[1:]
str_x = str_x[::-1]
x = '-' + str_x
x = int(x)
if x < -(2**31):
return 0
return x
# 解法2
def reverse(x):
if x == 0:
return 0
else:
str_x = str(x)
# 判断x为负数的情况
if str_x[0] == '-':
str_x = '-' + str_x[-1:-len(str_x):-1]
if int(str_x) < -(2**31):
return 0
else:
# 判断x为正数的情况
str_x = str_x[-1:- (len(str_x) + 1 ):-1]
if int(str_x) > 2**31 - 1:
return 0
return(int(str_x))
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