通过将长度至少为 K 的由偶数组成的子数组替换为其长度来减少给定数组

给定一个长度为N的数组arr[] ，如果长度大于或等于K ，任务是用它们的长度替换所有只有偶数元素的子数组。

例子：

Input: arr[] = {3, 6, 10, 2, 7, 6, 4, 8}, K = 2
Output: 3 3 7 3
Explanation: There are two subarrays with consecutive even numbers {6, 10, 2} and {6, 4, 8}. 
So they are replaced by their length 3 and 3 respectively.

Input: arr[] = {4, 8, 3, 6, 8}, K=3
Output: 4 8 3 6 8
Explanation: No subarray exists for which length of consecutive even numbers is greater than or equal to 3.

方法：按照以下步骤来回答这个问题：

• 创建两个向量，一个用于存储答案ans ，一个用于存储连续偶数temp 。
• 如果(K >= N) ，那么 返回原始向量arr 。
• 现在，遍历数组，并为每个元素：
  • 如果当前元素是奇数，则检查temp的长度是否大于或等于K 。
    • 如果是，则将其长度推入ans向量。
    • 否则将 temp 的元素推送到 ans向量中。
    • 清除临时向量。
  • 将当前元素推入ans向量中。
  • 如果当前元素是偶数，则将其推入temp 。
• 循环结束后，向量ans包含最终答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to replace all subarrays of
// even elements with their length if their
// length is greater than or equal to K
vector processArray(vector& arr,
                         int& K)
{
    int N = arr.size(), i, count = 0;
    vector ans;
    vector temp;
    if (K >= N)
        return arr;
 
    for (i = 0; i < N; i++) {
 
        // If arr[i] is odd
        if (arr[i] & 1) {
            if (temp.size() >= K)
                ans.push_back(temp.size());
            else {
                for (auto& x : temp) {
                    ans.push_back(x);
                }
            }
            ans.push_back(arr[i]);
            temp.clear();
        }
 
        // If arr[i] is even
        else {
            temp.push_back(arr[i]);
        }
    }
 
    if (temp.size() >= K) {
        ans.push_back(temp.size());
    }
    else {
        for (auto& x : temp) {
            ans.push_back(x);
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    vector arr
        = { 3, 6, 10, 2, 7, 6, 4, 8 };
    int K = 2;
    vector ans = processArray(arr, K);
 
    for (auto& x : ans)
        cout << x << " ";
 
    return 0;
}

Java

// Java program for the above approach
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
 
class GFG {
 
  // Function to replace all subarrays of
  // even elements with their length if their
  // length is greater than or equal to K
  static List processArray(List arr, int K) {
    int N = arr.size();
    ArrayList ans = new ArrayList();
    ArrayList temp = new ArrayList();
    ;
    if (K >= N)
      return arr;
 
    for (int i = 0; i < N; i++) {
 
      // If arr[i] is odd
      if ((arr.get(i) & 1) > 0) {
        if (temp.size() >= K)
          ans.add(temp.size());
        else {
          for (int x : temp) {
            ans.add(x);
          }
        }
        ans.add(arr.get(i));
        temp.clear();
      }
 
      // If arr[i] is even
      else {
        temp.add(arr.get(i));
      }
    }
 
    if (temp.size() >= K) {
      ans.add(temp.size());
    } else {
      for (int x : temp) {
        ans.add(x);
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void main(String args[]) {
    List arr = Arrays.asList(3, 6, 10, 2, 7, 6, 4, 8);
    int K = 2;
    List ans = processArray(arr, K);
 
    for (int x : ans)
      System.out.print(x + " ");
  }
}
 
// This code is contributed by saurabh_jaiswal.

Python3

# python3 program for the above approach
 
# Function to replace all subarrays of
# even elements with their length if their
# length is greater than or equal to K
def processArray(arr, K):
 
    N, count = len(arr), 0
    ans = []
    temp = []
    if (K >= N):
        return arr
 
    for i in range(0, N):
 
        # If arr[i] is odd
        if (arr[i] & 1):
            if (len(temp) >= K):
                ans.append(len(temp))
            else:
                for x in temp:
                    ans.append(x)
 
            ans.append(arr[i])
            temp.clear()
 
        # If arr[i] is even
        else:
            temp.append(arr[i])
 
    if (len(temp) >= K):
        ans.append(len(temp))
 
    else:
        for x in temp:
            ans.append(x)
 
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    arr = [3, 6, 10, 2, 7, 6, 4, 8]
    K = 2
    ans = processArray(arr, K)
 
    for x in ans:
        print(x, end=" ")
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG {
 
  // Function to replace all subarrays of
  // even elements with their length if their
  // length is greater than or equal to K
  static List processArray(List arr, int K) {
    int N = arr.Count;
    List ans = new List();
    List temp = new List();
    ;
    if (K >= N)
      return arr;
 
    for (int i = 0; i < N; i++) {
 
      // If arr[i] is odd
      if ((arr[i] & 1) > 0) {
        if (temp.Count >= K)
          ans.Add(temp.Count);
        else {
          foreach (int x in temp) {
            ans.Add(x);
          }
        }
        ans.Add(arr[i]);
        temp.Clear();
      }
 
      // If arr[i] is even
      else {
        temp.Add(arr[i]);
      }
    }
 
    if (temp.Count >= K) {
      ans.Add(temp.Count);
    } else {
      foreach (int x in temp) {
        ans.Add(x);
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void Main(String []args) {
    List arr = new List(new int[]{3, 6, 10, 2, 7, 6, 4, 8});
    int K = 2;
    List ans = processArray(arr, K);
 
    foreach (int x in ans)
      Console.Write(x + " ");
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

       // JavaScript code for the above approach
 
       // Function to replace all subarrays of
       // even elements with their length if their
       // length is greater than or equal to K
       function processArray(arr, K)
       {
           let N = arr.length, i, count = 0;
           let ans = [];
           let temp = [];
           if (K >= N)
               return arr;
 
           for (i = 0; i < N; i++) {
 
               // If arr[i] is odd
               if (arr[i] & 1) {
                   if (temp.length >= K)
                       ans.push(temp.length);
                   else {
                       for (let x of temp) {
                           ans.push(x);
                       }
                   }
                   ans.push(arr[i]);
                   temp = [];
               }
 
               // If arr[i] is even
               else {
                   temp.push(arr[i]);
               }
           }
 
           if (temp.length >= K) {
               ans.push(temp.length);
           }
           else {
               for (let x of temp) {
                   ans.push(x);
               }
           }
           return ans;
       }
 
       // Driver Code
 
       let arr
           = [3, 6, 10, 2, 7, 6, 4, 8];
       let K = 2;
       let ans = processArray(arr, K);
 
       for (let x of ans)
           document.write(x + " ");
      // This code is contributed by Potta Lokesh
   

输出

3 3 7 3 

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