Reverse digits of an integer.
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.
Example1: x = 123, return 321
Example2: x = -123, return -321
要求：反转整形数字，需要考虑到负数和整形数字溢出的问题
10100
=>101
1000000003
=>0
-2147483648
=>
2147483647
-2147483648
public int reverse(int x) { StringBuilder sb = new StringBuilder(); int mod, flag=0; boolean temp = false; if (x==0 || x==Integer.MIN_VALUE) return 0; if(x<0){ x = Math.abs(x); temp = true; } System.out.println(x); while(x>0){ mod = x%10; if(mod==0 && flag==0){}else{ sb.append(mod+""); flag++; } x = x/10; } long val = Long.valueOf(sb.toString()); int res = (val>Integer.MAX_VALUE)? 0:(int)val; if (temp) res = -res; return res; }
[LeetCode]-algorithms-Reverse Integer
原文：http://www.cnblogs.com/lianliang/p/5328382.html