Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

Solution: 注意overflow，还要Integer.MIN_VALUE和MAX VALUE值不同

    public int reverse(int x) {
        int sign = 1;
        if (x < 0) {
            sign = -1;
        }
        long temp = Math.abs((long)x), ans = 0;
        while (temp != 0) {
            ans = ans * 10 + temp % 10;
            if (ans > Integer.MAX_VALUE && sign == 1 || ans - 1 > Integer.MAX_VALUE && sign == -1) {
                return 0;
            }
            temp /= 10;
        }
        return (int)(sign * ans);
    }

其实可以直接result = (result*10) + (x%10)，不用取绝对值

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

    }

不用long: If overflow exists, the new result will not equal previous one.

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;
}