7. Reverse Integer

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本文介绍了一种有效的整数反转算法，解决了32位有符号整数的反转问题，并考虑了末尾为0的情况及反转可能导致的溢出问题。

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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?

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.

数字反转问题

首先复习一下：32位计算机字长,用于表示整数,共有2的32平方个.
所以,无符号整数的范围是0~2^32或0~4294967296
带符号整数,因为需要1位来表示+-,所以范围为
-2^31~2^31,或-2147483648~2147483648

每次对X取余就得到最后一个数字，然后除以10就去掉最后一个数字，先用long保存结果，判断是否超过上限/下限，如果超过就返回0.这样也不用考虑结尾是0 的情况。

public static int reverse(int x) {
		
		long res=0;
		while(x!=0){
			res = res*10+x%10;
			x = x/10;			
		}      
		if(res>Integer.MAX_VALUE||res<Integer.MIN_VALUE)
			return 0;
       return (int)res;
    }