[LeetCode] 087: Reverse Integer
2017-09-10 20:51
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[Problem]
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).
[Solution]
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).
[Solution]
class Solution { public: int reverse(int x) { // Note: The Solution object is instantiated only once and is reused by each test case. long long y = 0; bool mark = x < 0 ? true, x = -1*x : false; while(x > 0){ y = y*10 + x%10; x /= 10; } return mark ? -y : y; } };说明:版权所有,转载请注明出处。Coder007的博客
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