Algorithm_Number of Digit One(1的数目)

最新推荐文章于 2024-02-27 00:44:08 发布

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这篇博客探讨了如何计算在给定整数范围内数字1出现的次数，提供了两种不同的解题思路和方法。

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.

For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

给定一个整数n，计算所有小于等于n的非负整数中数字1出现的次数。

例如：

给定 n = 13，

返回6， 因为数字1在下面的数字中出现：1, 10, 11, 12, 13（共6次）。

解法一：

package com.liangdianshui;

import java.util.Scanner;

public class algorithm1 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long value = scanner.nextLong();
        long startTime = System.currentTimeMillis();
        String valueOf = String.valueOf(CountOne(value));
        long endTime = System.currentTimeMillis();
        long time = endTime - startTime;
        System.out.println("1的数目：" + valueOf + "\n用时：" + time+"ms");
    }

    public static long CountOne(long n) {
        long count = 0, i = 0, j = 1;
        for (i = 0; i <=n; i++) {
            j = i;
            while (j != 0) {
                if (j % 10 == 1) {
                    count++;
                }
                j /= 10;
            }
        }
        return count;
    }
}

解法二：

package com.liangdianshui;

import java.util.Scanner;

public class algorithm2 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long value = scanner.nextLong();
        long startTime = System.currentTimeMillis();
        String valueOf = String.valueOf(CountOne(value));
        long endTime = System.currentTimeMillis();
        long time = endTime - startTime;
        System.out.println("1的数目：" + valueOf + "\n用时：" + time+"ms");
    }

    public static long CountOne(long n) {
        long count = 0;
        if (n == 0)
            count = 0;
        else if (n > 1 && n < 10)
            count = 1;
        else {
            long highest = n;// 最高位的数字
            int bit = 0;
            while (highest >= 10) {
                highest = highest / 10;
                bit++;
            }

            int weight = (int) Math.pow(10, bit);// 最高位的权重，即最高位一个1代表的大小
            if (highest == 1) {
                count = CountOne(weight - 1) + CountOne(n - weight) + n
                        - weight + 1;
            } else {
                count = highest * CountOne(weight - 1)
                        + CountOne(n - highest * weight) + weight;
            }
        }
        return count;
    }
}

解法三：

package com.liangdianshui;

import java.util.Scanner;

public class algorithm3 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long value = scanner.nextLong();
        long startTime = System.currentTimeMillis();
        String valueOf = String.valueOf(CountOne(value));
        long endTime = System.currentTimeMillis();
        long time = endTime - startTime;
        System.out.println("1的数目：" + valueOf + "\n用时：" + time+"ms");
    }

    public static long CountOne(long n) {
        long count = 0;
        long i = 1;
        long current = 0, after = 0, before = 0;
        while ((n / i) != 0) {
            current = (n / i) % 10;
            before = n / (i * 10);
            after = n - (n / i) * i;

            if (current > 1)
                count = count + (before + 1) * i;
            else if (current == 0)
                count = count + before * i;
            else if (current == 1)
                count = count + before * i + after + 1;

            i = i * 10;
        }
        return count;

    }
}