Offer to prove safety fluent series of flow - a bit of a digital number sequence

A digital bit sequence of numbers: 44 face questions

One, Title Description

0123456789101112131415 ... in digital format into a sequence of character sequence. In this sequence, bit 5 (counting from 0) is 5, the 13th bit is 1, 4 bits are 19, and the like. Please write a function corresponding seeking any bit digital.

Second, the problem analysis

The search for effective solutions, but also find the law:

1. Digit number a total of 10, i.e. 0 to 9, * 1 accounted for 10 digits; (special)
2. A two-digit number a total of 90, i.e. 10 to 99, each figure represents two, accounting for a total of 90 * 2 digits;
3. ……
4. A total number m of bits of the 9 * 10 ^ (m-1) th, m bits representing each number, representing 9 * 10 ^ (m-1) * m bit number.

Analyzing the n-th number belongs to a few, then to find the number of several.

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For example:
0123456789101112131415161718192021 ...

We want to take the first 33.

For more images, we put the law above abstract scene into a more intuitive: we want to put actual figures are stored in a squares inside. A plurality of digital grid may be stored.

First of all we want to find the first 33-digit number that corresponds to the character. Since the number of digits are 10, 90 have a double-digit number, 33 interposed between 10-100, two may be determined.

A storage box using two features:
(33-10) / 2 stars 11, 10 will come together with figure 33 corresponding to the character of the number is 21
(33-10) we come to 2% corresponding character bit, i.e., a first digital

Third, questions

  public int digitAtIndex(int index) {
        if(index<0){
            return -1;
        }
        // m位数
        int m=1;
        while(true) {
            // m位数的个数
            int numbers=numbersOfIntegers(m);
            if(index<numbers*m) {
                return getDigit(index,m);
            }
            index-=numbers*m;
            m++;
        }
    }

    // 返回m位数的总个数
    private int numbersOfIntegers(int m) {
        if(m == 1) {
            return 10;
        }
        return (int) (9*Math.pow(10, m-1));
    }

    // 获取数字
    private int getDigit(int index, int m) {
        // 对应的m位数
        int number = getFirstNumber(m)+index/m;
        // 在数字中的位置
        int indexFromRight = m - index%m;
        for(int i=1; i < indexFromRight; i++) {
            number/=10;
        }
        return number%10;
    }

    // 第一个m位数 例如第一个两位数是10，第一个三位数是100
    private int getFirstNumber(int m) {
        if(m==1) {
            return 0;
        }
        return (int) Math.pow(10, m-1);
    }