233. Number of Digit One *HARD* -- 从1到n的整数中数字1出现的次数

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.

Hint:

1. Beware of overflow.

class Solution {
public:
    int countDigitOne(int n) {
        if(n <= )
            return ;
        vector<int> v;
        int t = n;
        while(t)
        {
            v.push_back(t%);
            t /= ;
        }
        int l = v.size(), rest = n - v[l-] * pow(, l-);
        return (v[l-] >  ? pow(, l-) : rest + ) + v[l-] * (l-) * pow(, l-) + countDigitOne(rest);
    }
};