Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter)
leetcode的测试样例中似乎没有翻转后溢出的情况。 本文地址
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class
Solution {
public
:
int
reverse(
int
x) {
bool
isPositive =
true
;
if
(x < 0){isPositive =
false
; x *= -1;}
long
long
res = 0;
//为了防止溢出，用long long
while
(x)
{
res = res*10 + x%10;
x /= 10;
}
if
(res > INT_MAX)
return
isPositive ? INT_MAX : INT_MIN;
if
(!isPositive)
return
res*-1;
else
return
res;
}
};
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本文转自tenos博客园博客，原文链接：http://www.cnblogs.com/TenosDoIt/p/3735866.html，如需转载请自行联系原作者
