[Offer] [17] [n print to the maximum number of bits 1]

• Title Description
• Ideas analysis
• Test Case
• Java code
• Code link

Title Description

　　Enter the number n, in order to print out a decimal from 1 to the maximum n-bit number. For example, input 3, then print out up to a maximum of 1, 2, 3 digits, ie 999.

Ideas analysis

1. To take into account the problem of large numbers, we ask whether the n-digit integer, there is no obvious limitations of size n, we can use the string to print the number of bits n
2. We need to do two things:
   • In the digital representation of the analog +1 string, the string plus one operation, add a note of the time is stopped, i.e., when the most significant bit + 1, if a carry is stopped plus 1, for example: 1 + 999
   • n print string representation of the number, digital printing character array formation, not the time to pay attention to when faced with '0' to start printing

Test Case

1. Function test (input 1,2,3 ......)
2. Special test input (input 0, -1)

Java code

public class Offer17 {

    public static void main(String[] args) {
        System.out.println("功能测试--->");
        test1();
        System.out.println("特殊值测试--->");
        test2();
    }

    public static void print1ToMaxOfNDigits(int n) {
        Solution1(n);
    }

    /**
     * 用模拟加1 的方法
     * 
     * @param n
     */
    private static void Solution1(int n) {
        if (n <= 0) {
            throw new IllegalArgumentException("参数非法");
        }
        char[] number = new char[n];
        Arrays.fill(number, '0');// 全部初始化为为'0'
        while (!increment(number)) {
            printCharNumber(number);
        }
    }

    /**
     * 对字符串进行加一操作,注意停止加一的时机，即当最高位+1时，如果产生进位，则停止加1 ，比如：999+1
     * 
     * @param number
     * @return
     */
    private static boolean increment(char[] number) {
        boolean isOverflow = false; // 判断最高位是否产生进位，停止+1的条件
        int nTakeOver = 0; // 产生的进位
        // 数组从后往前遍历，数组后面的位表示数值的低位
        for (int i = number.length - 1; i >= 0; i--) {
            int nSum = (number[i] - '0') + nTakeOver;
            if (i == number.length - 1) {
                nSum++;
            }
            if (nSum >= 10) {
                if (i == 0) {// 最高位
                    isOverflow = true;
                } else {
                    nSum -= 10;
                    nTakeOver = 1;
                    number[i] = (char) ('0' + nSum);
                }
            } else {
                number[i] = (char) ('0' + nSum);
                break;
            }
        }
        return isOverflow;
    }

    /**
     * 打印字符数组形成的数字，注意当遇到不是'0'的时候才开始打印
     */
    private static void printCharNumber(char[] number) {
        boolean isBeginning0 = true;
        for (int i = 0; i < number.length; i++) {
            if (isBeginning0 && (number[i] - '0') != 0) {
                isBeginning0 = false;
            }
            if (!isBeginning0) {
                System.out.print(number[i]);
            }
        }
        System.out.println();
    }

    private static void test1() {
        System.out.println("2---->");
        print1ToMaxOfNDigits(2);
    }

    private static void test2() {
        System.out.println("-1---->");
        print1ToMaxOfNDigits(-1);
        System.out.println("0---->");
        print1ToMaxOfNDigits(0);
    }
}

Code link

Offer to prove safety codes -Java