[LeetCode] 233 number, the number 1

Title Description

Given an integer n, is less than the calculated number of all non-negative integer equal to n the number 1 appears.

输入: 13
输出: 6 
解释: 数字 1 出现在以下数字中: 1, 10, 11, 12, 13 。

Reference Code

"Prove safety offer" 43 questions, see the book solution.
From the "digital laws" to proceed, step by step to find out the law through specific examples.

C ++ language version

class Solution {
public:
    int countDigitOne(int n){
        if(n <= 0)
            return 0;
        
        string str = to_string(n);
        return mySolutionRecursively(str);
    }
    
    int mySolutionRecursively(string str){  // eg: 21345
        int length = str.size();
        if(length == 1){
            if(str[0] == '0')
                return 0;
            else
                return 1;
        }
        
        int firstNum = str[0] - '0';
        int amountOfFirstDigit = 0;
        if(firstNum == 1)  // 当firstNum == 0时，此时amountOfFirstDigit == 0
            amountOfFirstDigit = stoi(str.substr(1)) + 1;
        else if(firstNum > 1)
            amountOfFirstDigit = (int)pow(10, length-1);
        
        int amountOfOtherDigit = 0;
        amountOfOtherDigit = firstNum * (length-1) * (int)pow(10, length-2);
        
        return amountOfFirstDigit + amountOfOtherDigit + mySolutionRecursively(str.substr(1));
    }
    
};

C language version

class Solution {
public:
    int countDigitOne(int n)
    {
        if(n <= 0)
            return 0;
        
        char strN[50];
        sprintf(strN, "%d", n);
        
        return mySolutionRecursively(strN);
    }
    
    int mySolutionRecursively(char *str){
        int length = strlen(str);
        if(length == 1){
            if(str[0] == '0')
                return 0;
            else
                return 1;
        }
        
        int firstNum = str[0] - '0';
        int amountOfFirstDigit = 0;
        if(firstNum == 1)
            amountOfFirstDigit = (int)atoi(str+1) + 1;
        else if(firstNum > 1)
            amountOfFirstDigit = (int)pow(10, length-1);
        
        int amountOfOtherDigit = 0;
        amountOfOtherDigit = firstNum * (length-1) * (int)pow(10, length-2);
        
        return amountOfFirstDigit + amountOfOtherDigit + mySolutionRecursively(str+1);
    }
    
};