class Solution {
public:
    int countDigitOne(int n) {
        if(!n) return 0;
        vector<int> number;
        while(n) number.push_back(n % 10), n /= 10;
        long long res = 0;
        for(int i = number.size() - 1; i >= 0; i--)
        {
            int left = 0, right = 0, t = 1;

            for(int j = number.size() - 1; j > i; j-- )
                left = left * 10 + number[j];

            for(int j = i - 1; j >= 0; j --)
                right = right * 10 + number[j], t *= 10;

            res += left * t;
            if(number[i] == 1) res += right + 1;
            else if(number[i] > 1) res += t;
        }
        return res;
    }
};

【剑指offer】【其他算法】43. 1～n整数中1出现的次数

原文：https://www.cnblogs.com/Trevo/p/12823056.html