题意：给定一个整数n，打印出从1到最大n位数。
思路：一般解法很简单，但是基于效率考虑，选择位运算来解答。
代码：

package MianShiTi_12;

public class MianShiTi_12 {
    public static  boolean isOverflow(int[] number) throws Exception {
        if(number.length < 1){
            throw new Exception("invalid length of number");
        }
        boolean isOverflow = false;
        //进位
        int carry = 0;
        for(int i = number.length - 1; i >= 0 ;i--){
            int sum = number[i] + carry;
            if(i == number.length - 1){
                sum++;
            }
            if(sum >9 ){
                if(i == 0){
                    isOverflow = true;
                }else{
                    carry = 1;
                    number[i] = 0;
                    sum = 0;
                }
            }else{
                number[i] = sum;
                break;
            }
        }
        return isOverflow;
    }

    public static void print(int[] number) {
        boolean isBegining = true;
        for(int i = 0; i< number.length ; i++){
            if(isBegining && number[i] != 0){
                isBegining = false;
            }
            if(!isBegining){
                System.out.println(number[i]);
            }
        }
    }

    public static void main(String [] args) throws Exception {
        int[] number = new int[3];
        while(!isOverflow(number)){
            print(number);
            System.out.println();
        }
    }
}