Reverse digits of an integer.
Example1: x = 123, return 321 Example2: x = -123, return -321
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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.
Note: The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
主要的问题就是溢出，所以就偷懒用了 long 型保存中间数据
public class Solution { public int reverse(int x) { long ans = 0; long temp = Math.abs((long)x); while(temp!=0){ ans = 10 * ans + temp % 10; temp = temp / 10; } if(x<0){ ans = -ans; if(ans<Integer.MIN_VALUE) return 0; }else{ if(ans>Integer.MAX_VALUE) return 0; } return (int)ans; } } 好像负数也不需要转成正数然后反转，直接就可以。看了别人的方法，在溢出的判断上不需要这么麻烦，反过来与前一个值判断即可
public int reverse(int x) { int result = 0; while (x != 0) { int tail = x % 10; int newResult = result * 10 + tail; if ((newResult - tail) / 10 != result) { return 0; } result = newResult; x = x / 10; } return result; }