LeetCode "Number of Digit One"

Marked as mathematics problem, but if you are able to divide 1s into several cases, the AC code becomes natural.

class Solution
{
unordered_map<long long, int> memo;
public:
int countDigitOne(int n)
{
if (n < 10) return n < 1 ? 0 : 1;
if (memo.find(n) != memo.end())    return memo
;

int base = pow(10, floor(log10(n)));
int v = n / base;
int r = n % base;

int cnt0 = countDigitOne(base - 1);
int cnt1 = 0;
if (v == 1)
{
cnt1 = r + 1;
}
else
{
cnt1 = base + cnt0;
cnt1 += (v - 2) * cnt0;
}

int ret = cnt0 + cnt1 + countDigitOne(r);
memo
= ret;
return ret;
}
};

Lesson learnt: visualization in mind is always helpful.

And this link is amazing: https://leetcode.com/discuss/44281/4-lines-o-log-n-c-java-python - Count it digit by digit.