233. Number of Digit One

题目：

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

链接： http://leetcode.com/problems/number-of-digit-one/

题解：

又是数学题，主要思路是用递归来做。还有一些别的解法，二刷的时候争取理解最优解。

下面我们来一步步分析：

1. n < 1时，结果为0
2. 1 <= n < 10时，结果为1
假定n = 312，我们把这个计算过程分解为几个步骤:
1. (1 ~ 99)， 结果为 countDigitOne(99)
2. (100 ~ 199)， 结果为 100 + countDigitOne(99)
3. (200 ~ 299)， 结果为countDigitOne(99)
4. (300 ~ 312)， 结果为countDigitOne(12)
3. 假定n = 112， 我们也把这个计算过程分解一下:
   1. (1 ~ 99)， 结果为 countDigitOne(99)
   2. (100 ~ 112)， 结果为 112 - 100 + 1 + countDigitOne(12)
4. 由此我们可以推出通项公式

Time Complexity  - O(log₁₀n)， Space Complexity - O(log₁₀n)

public class Solution {
    public int countDigitOne(int n) {
        if (n < 1)
            return 0;
        if (n < 10)
            return 1;
        int baseInTen = (int)Math.pow(10, String.valueOf(n).length() - 1);   
        int highestDigit = n / baseInTen;         // get the highest digit of n
        
        if(highestDigit == 1)
            return countDigitOne(baseInTen - 1) + (n - baseInTen + 1) + countDigitOne(n % baseInTen);
        else
            return highestDigit * countDigitOne(baseInTen - 1) + baseInTen + countDigitOne(n % baseInTen);
    }
}

Reference:

https://leetcode.com/discuss/44281/4-lines-o-log-n-c-java-python

https://leetcode.com/discuss/44279/clean-c-code-of-log10-complexity-with-detailed-explanation

https://leetcode.com/discuss/44314/accepted-solution-using-counting-principle-with-explanation

https://leetcode.com/discuss/44465/my-ac-java-solution-with-explanation

https://leetcode.com/discuss/44617/my-recursion-implementation

https://leetcode.com/discuss/47774/0ms-recursive-solution-in-c-8-line-code

https://leetcode.com/discuss/46366/ac-short-java-solution

https://leetcode.com/discuss/64604/my-simple-and-understandable-java-solution

https://leetcode.com/discuss/64962/java-python-one-pass-solution-easy-to-understand

https://leetcode.com/discuss/54107/0-ms-recursive-solution

https://leetcode.com/discuss/58868/easy-understand-java-solution-with-detailed-explaination