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Number of Digit One

时间:2016-01-19 12:02:08      阅读:149      评论:0      收藏:0      [点我收藏+]

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.

public class Solution {
    public int countDigitOne(int n) {
      int multiply = 1;
        int sum = 0;
        int input = n;

        int stub = 0;
        while (n > 0) {
            if (n % 10 > 0) {
                stub = n / 10 + 1;
                sum += (n / 10 + 1) * multiply;
            } else {
                stub = n / 10;
                sum += (n / 10) * multiply;
            }

            if (n % 10 == 1 && n / 10 > 0) {
                sum -= (stub * multiply - input % (multiply * 10) - 1)
                        % multiply;
            } else if (n % 10 == 1 && n / 10 == 0) {
                sum -= (stub * multiply - input % (multiply) - 1) % multiply;
            }

            multiply = multiply * 10;
            n = n / 10;
        }
        return sum;
    }
}

 

Number of Digit One

原文:http://www.cnblogs.com/fgynew/p/5141495.html

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