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.

本以为比较难，实际写起来还可以。按位考虑出现的次数，个位，十位，百位。每位出现1的次数为高位代表的数字乘以当前位的大小，然后考虑一下边界条件，当前位小于或者正好等于1的情况。

class Solution {public:    int countDigitOne(int n) {        if (n <= 0) {            return 0;        }                int counts = 0;        int lowUnit = 1;        int lowNum = 0;        while (n > 0) {            int highNum = n / 10;            counts += highNum * lowUnit;            if (n % 10 > 1) {                counts += lowUnit;            }            else if (n % 10 == 1) {                counts += lowNum + 1;            }                        lowNum += (n % 10) * lowUnit;            n /= 10;            lowUnit *= 10;        }                return counts;    }};