《剑指offer》面试题17：打印从1到最大的n位数

面试题17：打印从1到最大的n位数

代码：

public class Print1toN {
    //输进参数N，打印从1到最大的N位十进制数
    public static void print1toMaxOfNDigits(int n) {
        if (n<=0) {
            System.out.println("参数必须是非负正整数");
            return;
        }
        char [] numbers = new char[n];
        //初始化字符数组
        for (int i = 0; i < numbers.length; i++) {
            numbers[i] = '0';
        }
        while (!arrayOutAfterAdd(numbers)) {
            printNumber(numbers);
        }
        //System.out.println(numbers[n-1]);
    }
    private static void printNumber(char[] numbers) {
        int j  = 0;
        while (numbers[j] =='0') {
            j++;
        }
        for (int i = j; i < numbers.length; i++) {
            
            System.out.print(numbers[i]);
        }
        System.out.println(" ");
    }
    private static boolean arrayOutAfterAdd(char[] numbers) {
        // 当前字符数组加1后最前一位不会进位，返回true
        boolean isOverFlow = false;
        int nTakeOver = 0;
        int length = numbers.length;
        int nSum = 0;
        for (int i = length-1; i >=0; i--) {
            nSum = numbers[i]-'0' + nTakeOver;
            if(i == length-1) {
                nSum ++;
            }
            if (nSum >=10) {
                if (i == 0) {
                    isOverFlow = true;
                    
                }else{
                    nTakeOver = 1;
                    nSum -=10;
                    numbers[i] = (char) ('0'+nSum);
                }
            }else{
                numbers[i] = (char)('0'+nSum);
                //既然不用进位，那么就可以跳出循环了，因为后面的位的数字均不用改变
                break;
            }
            
        }
        return isOverFlow;
    }
    public static void main(String[] args) {
        print1toMaxOfNDigits(3);
    }
}