从1到n整数中1出现的个数

参考了力扣区路飞大佬的题解：

public class Solution {
    public int NumberOf1Between1AndN_Solution(int n) {
        int digit = 1, res = 0;
        int high = n / 10, cur = n % 10, low = 0;
        while(high != 0 || cur != 0){
            if(cur == 0)
                res += high * digit;
            else if(cur == 1)
                res += high * digit + low + 1;
            else res += (high + 1) * digit;
            low += cur * digit;
            cur = high % 10;
            high /= 10;
            digit *= 10;
        }
        return res;
    }
}

时间复杂度： 数字N的位数： O(log10N) = O(logN)
空间复杂度： O(1)