題目：

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.

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

題解：

又是數學題，主要思路是用遞歸來做。還有一些別的解法，二刷的時候爭取理解最優解。

下面我們來一步步分析：

n < 1時，結果為0

1 <= n < 10時，結果為1 假定n = 312，我們把這個計算過程分解為幾個步驟:

(1 ~ 99)， 結果為 countDigitOne(99)

(100 ~ 199)， 結果為 100 + countDigitOne(99)

(200 ~ 299)， 結果為countDigitOne(99)

(300 ~ 312)， 結果為countDigitOne(12)

假定n = 112， 我們也把這個計算過程分解一下:

(1 ~ 99)， 結果為?countDigitOne(99)

(100 ~ 112)， 結果為 112 - 100 + 1 + countDigitOne(12)

由此我們可以推出通項公式

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 nif(highestDigit == 1)return countDigitOne(baseInTen - 1) + (n - baseInTen + 1) + countDigitOne(n % baseInTen);elsereturn 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

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