leetcode：7. Reverse Integer

这题简单，也花了我好长时间，我自己写的code比较麻烦，也没啥技巧：按正负性分类执行，先转化成字符串，用stringbuilder进行旋转，如果超出范围了就用try catch

public int reverse(int x) {
        try {
            int result = 0;
            if (x == 0) {
                return x;
            } else if (x < 0) {
                String num = Integer.toString(0 - x);
                String newNum = new StringBuilder(num).reverse().toString();
                result = 0 - Integer.parseInt(newNum);
            } else {
                String num = Integer.toString(x);
                String newNum = new StringBuilder(num).reverse().toString();
                result = Integer.parseInt(newNum);
            }

            return result;
        } catch (Exception e) {
            return 0;
        }

    }

这种方法要好一些：https://discuss.leetcode.com/topic/15134/very-short-7-lines-and-elegant-solution

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