LeetCode233 Number of Digit One

计数数字中1的出现次数

最新推荐文章于 2021-08-13 01:56:42 发布

原创最新推荐文章于 2021-08-13 01:56:42 发布 · 213 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LeetCode

LeetCode 专栏收录该内容

53 篇文章

订阅专栏

本文介绍了一个算法问题，即计算从0到n的所有非负整数中数字1出现的总次数。通过一个Java实现的例子，展示了如何高效地解决这个问题。示例中包括了对13作为输入的解释，以及一个对大规模数字1410065408进行计算的演示。

题目：

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

Example:

Input: 13

Output: 6

Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

public class NumberDigitOne {
    public int countDigitOne(int n) {
        int count = 0, previous = 0, coef = 1;
        while (n > 0) {
            int remain = n % 10;
            int over = n / 10;
            if (remain > 1) {
                count += coef;
            } else if (remain == 1) {
                count += previous + 1;
            }
            count += coef * over;
            previous += coef * remain;
            coef *= 10;
            n /= 10;
        }
        return count;
    }

    public static void main(String[] args) {
        int n = 1410065408;
        System.out.println(new NumberDigitOne().countDigitOne(n));
    }
}