leetcode:Number of Digit One

最新推荐文章于 2021-08-13 01:56:42 发布

martin_liang

最新推荐文章于 2021-08-13 01:56:42 发布

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本文提供了一种算法来计算小于等于n的所有非负整数中1的总数量，通过分治策略遍历每个位数进行计算。

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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.

class Solution {
    
private:
    int countDigit(int n, int digit)
    {
        long num = n;
        int count = 0;
        long curDigitPos = 1;
   
        while (num/curDigitPos)
        {
            int currentDigit = (num%(curDigitPos*10))/curDigitPos;
            int upperDigits = num/(curDigitPos*10);
      
            if (currentDigit > digit)
            {
                count += (upperDigits+1) * curDigitPos;
            }
            else if (currentDigit < digit)
            {
                count += upperDigits * curDigitPos;
            }
            else
            {
                count += upperDigits * curDigitPos + num % curDigitPos + 1;
            }
      
            curDigitPos = curDigitPos * 10;
        }
   
        return count;
    }
    
public:
    int countDigitOne(int n) {
        if (n <= 0)
            return 0;
        
        return countDigit(n, 1);
    }
};