Number of Digit One

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，用于计算给定整数n下，所有小于等于n的非负整数中数字1出现的总次数。

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 {
public:
    int countDigitOne(int n) {
        if (n <= 0) 
        {
            return 0;
        }
        int curr = n, right = 0, result = 0;
        while (curr > 0)
        {
            if (curr%10 > 1)
            {
                result += (curr/10+1)*pow(10, right);
            }
            else if (curr%10 == 1)
            {
                result += (curr/10)*pow(10, right) + n%int(pow(10, right)) + 1;
            }
            else
            {
                result += (curr/10)*pow(10, right);
            }
            curr /= 10; 
            right++;
        }
        return result;
    }
};