LeetCode-7. Reverse Integer
2016-04-17 21:19
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Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
click to show spoilers.
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?
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Update (2014-11-10):
Test cases had been added to test the overflow behavior.
这道题就难在判断溢出这个地方,这里我的想法是转换为字符串,逆向转换字符串,再转回来,有异常就返回0
Example1: x = 123, return 321
Example2: x = -123, return -321
click to show spoilers.
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?
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Update (2014-11-10):
Test cases had been added to test the overflow behavior.
这道题就难在判断溢出这个地方,这里我的想法是转换为字符串,逆向转换字符串,再转回来,有异常就返回0
public class Solution { public int reverse(int x) { boolean flag = true; if(x < 0){ x = -x; flag = false; } String num = Integer.toString(x); String rev =""; for(int i = num.length()-1; i>=0; i--){ rev += num.charAt(i); } int ans = 0; try{ ans=Integer.parseInt(rev); }catch(Exception e){ ans = 0; } if(flag == true ){ return ans; }else{ return -ans; } } }
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