到 N 的最大数字集，使得 i 或 i/2 存在

给定一个正整数N ，任务是找到可以生成的最大集合的长度，使得如果i存在，则i/2将不存在并且1<=i<=N 。

Note: For multiple solutions, print anyone satisfying the condition.

例子：

Input: N = 2
Output: 1
Explanation: There are two possible values 1 and 2. If 2 is present, 1 can not be present since 2/2=1. So only possibilities are 1 or 2. Both 1 and 2 separately can be the answers.

Input: N = 10
Output: 1 3 4 5 7 9
Explanation: In the output, there are no i for which i/2 exists.

编程需要懂一点英语

方法：创建一个地图以便于存储和搜索，并创建一个向量来存储最终集。运行从1到N的循环（比如i ）如果i是奇数，则将i添加到答案向量并作为地图中的键。否则，如果i是偶数，则在地图中搜索i/2作为键。如果地图中不存在i/2 ，则将i添加到答案向量并作为地图中的键。请按照以下步骤解决问题：

初始化 map mp[]。
初始化 vector ans[]以存储结果。
使用变量i遍历范围[, N]并执行以下任务：
- 如果i是奇数，则将i推入向量ans[]并将{i, 1}插入映射mp[]。
- 否则，如果映射mp[]中不存在i/2 ，则将i推入向量ans[]并将{i, 1}插入映射mp[]。
执行上述步骤后，打印向量ans[]的值作为答案。

下面是上述方法的实现。

C++14

// C++ program for the above approach
#include 
using namespace std;
 
// Function to get the largest set
vector reqsubarray(int& n)
{
 
    // Initialize the map
    unordered_map mp;
 
    // Vector to store the answer
    vector ans;
    int i;
 
    // Traverse the range
    for (i = 1; i <= n; i++) {
        if (i & 1) {
            ans.push_back(i);
            mp.insert({ i, 1 });
        }
        else if (!mp[i/2]) {
            ans.push_back(i);
            mp.insert({ i, 1 });
        }
    }
 
    // Return the answer
    return ans;
}
 
// Driver Code
int main()
{
    int n = 10, i;
 
    vector ans;
 
    ans = reqsubarray(n);
 
    for (i = 0; i < ans.size(); i++)
        cout << ans[i] << " ";
 
    return 0;
}

Java

// Java program for the above approach
import java.util.ArrayList;
import java.util.HashMap;
 
class GFG
{
 
  // Function to get the largest set
  static ArrayList reqsubarray(int n)
  {
 
    // Initialize the map
    HashMap mp = new HashMap();
 
    // Vector to store the answer
    ArrayList ans = new ArrayList();
    int i;
 
    // Traverse the range
    for (i = 1; i <= n; i++) {
      if ((i & 1) == 1) {
        ans.add(i);
        mp.put(i, 1);
      }
      else if (!mp.containsKey(i / 2)) {
        ans.add(i);
        mp.put(i, 1);
      }
    }
 
    // Return the answer
    return ans;
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int n = 10, i;
 
    ArrayList ans = reqsubarray(n);
 
    for (i = 0; i < ans.size(); i++)
      System.out.print(ans.get(i) + " ");
 
  }
}
 
// This code is contributed by Saurabh Jaiswal

Python3

# Python 3 program for the above approach
 
# Function to get the largest set
def reqsubarray(n):
 
    # Initialize the map
    mp = {}
 
    # Vector to store the answer
    ans = []
 
    # Traverse the range
    for i in range(1, n + 1):
        if (i & 1):
            ans.append(i)
            mp[i]= 1
 
        elif ((i // 2) not in mp):
            ans.append(i)
            mp[i] = 1
 
    # Return the answer
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    n = 10
 
    ans = reqsubarray(n)
 
    for i in range(len(ans)):
        print(ans[i], end=" ")
 
        # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
 
  // Function to get the largest set
  static ArrayList reqsubarray(int n)
  {
 
    // Initialize the map
    Dictionary mp =
      new Dictionary();
 
    // Vector to store the answer
    ArrayList ans = new ArrayList();
    int i;
 
    // Traverse the range
    for (i = 1; i <= n; i++) {
      if ((i & 1) == 1) {
        ans.Add(i);
        mp.Add(i, 1);
      }
      else if (!mp.ContainsKey(i / 2)) {
        ans.Add(i);
        mp.Add(i, 1);
      }
    }
 
    // Return the answer
    return ans;
  }
 
  // Driver Code
  public static void Main()
  {
    int n = 10, i;
 
    ArrayList ans = reqsubarray(n);
 
    for (i = 0; i < ans.Count; i++)
      Console.Write(ans[i] + " ");
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

1 3 4 5 7 9

时间复杂度： O(N)
辅助空间： O(N)