Leetcode: Number of Digit One

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:

Beware of overflow.

this one is the same with Lintcode: Digit Counts, 小心溢出，所以用了long型

当某一位的数字小于i时，那么该位出现i的次数为：更高位数字x当前位数
当某一位的数字等于i时，那么该位出现i的次数为：更高位数字x当前位数+低位数字+1
当某一位的数字大于i时，那么该位出现i的次数为：(更高位数字+1)x当前位数

public class Solution {
    public int countDigitOne(int n) {
        //I will count how many 1s appear on each bit of n, respectively.
        if (n <= 0) return 0;
        long num = n;
        long bit = 1;
        int res = 0;
        while (num/bit > 0) {
            long upper = num/(bit*10);
            long cur = num%(bit*10)/bit;
            long lower = num%bit;
            
            if (cur < 1) 
                res += upper*bit;
            else if (cur == 1) 
                res += upper*bit + lower + 1;
            else
                res += (upper+1) * bit;
                
            bit *= 10;
        }
        
        return res;
    }
}