统计桌面上的不同数字

You are given a positive integer n, that is initially placed on a board. Every day, for 109 days, you perform the following procedure:

• For each number x present on the board, find all numbers 1 <= i <= n such that x % i == 1.
• Then, place those numbers on the board.

Return the number of distinct integers present on the board after 109 days have elapsed.

Note:

• Once a number is placed on the board, it will remain on it until the end.
• % stands for the modulo operation. For example, 14 % 3 is 2.

 

Example 1:

Input: n = 5
Output: 4
Explanation: Initially, 5 is present on the board. 
The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1. 
After that day, 3 will be added to the board because 4 % 3 == 1. 
At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5. 

Example 2:

Input: n = 3
Output: 2
Explanation: 
Since 3 % 2 == 1, 2 will be added to the board. 
After a billion days, the only two distinct numbers on the board are 2 and 3. 

 

Constraints:

• 1 <= n <= 100

/*
 * 题目思路：
 * 与Java普通解法类似，但是利用Java Stream流式编程来实现。
 * 初始时数字n在桌面上，之后每一天都会在桌面上寻找满足条件的数字。
 * 这些满足条件的数字会一直保留在桌面上直到结束。
 * 因此，我们只需要计算出这些满足条件的数字数量即可。
 */
import java.util.stream.IntStream;

public class Solution {
    public int distinctIntegers(int n) {
        // 初始值为1，因为n本身就在桌面上
        return (int) IntStream.range(2, n) // 生成范围为2到n-1的数字流
                .filter(i -> n % i == 1) // 过滤出n % i == 1的数字
                .count() + 1; // 计数，并加上n本身的1个计数
    }
}

此题解使用一个集合 `board` 来存储每一天结束时桌面上的所有数字。初始时，集合中只包含数字 `n`。每天对当前集合中的每个数字 `x` 进行遍历，对于每个 `x`，再次遍历从 `1` 到 `x-1` 的所有整数 `i`，检查 `x % i == 1` 是否成立。如果条件满足，则将 `i` 添加到一个新的临时集合 `new_board` 中。如果在某一天结束时 `new_board` 和 `board` 相同，意味着没有新的数字被添加到集合中，因此可以停止迭代，返回集合中不同整数的数目。这反映出了算法的收敛性，即达到了一个固定点，新的数字不再被生成。

O(n^2)

O(n)