Reverse Integer–LeetCode#7

1.Reverse Integer

Reverse digits of an integer.

(反转整形数字)

Example:

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.(输入为32位整数，直接返回0)

思路：直接转化为字符串数组循环遍历，通过StringBuilder连接。

public class ReverseInteger {
    public int reverse(int x) {
        boolean flag = false;
        int result = 0;
        if (x < 0) {
            flag = true;
            x = Math.abs(x);
        } else if (x == 0) {
            return 0;
        }
        String str = String.valueOf(x);
        char[] chars = str.toCharArray();
        StringBuilder stringBuilder = new StringBuilder();
        for (int i = str.length(); i > 0; i--) {
            stringBuilder.append(chars[i - 1]);
        }
        try {
            if (flag) {
                result = -(Integer.valueOf(stringBuilder.toString()));
            } else {
                result = Integer.valueOf(stringBuilder.toString());
            }
        } catch (NumberFormatException e) {
            return 0;
        }
        return result;
    }
}

这种方法通过转化成char数字来操作x中每个数字(显然非常的low).
可以用递推算法实现，int类型上通过 %10 求最后一位数字。

public class ReverseInteger {
    public int reverse(int x) {
        long result = 0;
        while(x != 0)
        {
            result = (result*10) + (x%10);
            if(result > Integer.MAX_VALUE) return 0;
            if(result < Integer.MIN_VALUE) return 0;
            x = x/10;
        }
        return (int)result;
    }
    public static void main(String[] args) {
        ReverseInteger reverseInteger = new ReverseInteger();
        System.out.println(reverseInteger.reverse(123));
    }
}