给定一个包含N个数字的数组arr [] ，任务是通过用该子数组的第一个偶数元素替换所有连续的偶数来合并连续偶数的子数组。
注意：如果在给定序列中至少有三个偶数，则一系列偶数整数被认为是连续的。因此，在子数组中合并具有至少3个连续偶数的元素。
例子：
Input: arr[] = {2, 2, 2, 100, 5, 4, 2, 9, 10, 88, 24}
Output: 2 5 4 2 9 10
Explanation:
The given series contains two consecutive even subsequences. They are:
{2, 2, 2, 100}, {10, 88, 24}. The two subsequences have to be merged to the first element in the subseries.
Therefore, {2, 2, 2, 100} is replaced by 2 and {10, 88, 24} is replaced by 10 in the original series.
Input: arr[] = {2, 4, 5, 3, 6, 8, 10, 3, 4}
Output: 2 4 5 3 6 3 4
方法：为了解决此问题，我们首先需要确定是否存在大小大于3的连续偶数子序列。因此，想法是遍历给定的数组并检查数字是否为偶数。
- 遍历数组。
- 检查元素是否为偶数。
- 如果是偶数，则创建一个临时数组来存储下一个连续的偶数，直到下一个数为奇数为止。
- 继续在临时数组中添加元素，直到出现奇数为止。
- 如果此临时数组的大小大于3，则从给定数组中删除这些元素，并用该临时数组的第一个元素替换它们。
- 清空临时数组以计算下一组偶数子序列。
下面是上述方法的实现：
C++
// C++ program to merge the array
// as per the given condition
#include
using namespace std;
// Function to merge the array
// as per the given condition
vector merge(vector arr)
{
// Variable to store the final
// sequence
vector ans;
// Temporary array to store the
// even numbers while traversing
vector e;
int i = 0;
int j = 0;
int count = 0;
// Iterating through the array
while(i < arr.size())
{
// If the element is even
if(arr[i] % 2 == 0)
{
j = i;
// Iterating till an odd element
// is found
while(j < arr.size())
{
// Keep appending into the
// temporary array if the
// even number is occurred
if (arr[j] % 2 == 0)
{
e.push_back(arr[j]);
count += 1;
}
// Break if an odd number
// has occurred
else
break;
j += 1;
}
// If the series has at least
// three elements, then merge
if(count >= 3)
ans.push_back(e[0]);
// Else, add all elements to the
// answer array
else
{
for (auto i: e)
ans.push_back(i);
}
// Reseting the count and
// temp array
count = 0;
e.clear();
i = j;
}
// If the element is odd, add
// it to the answer array
else
{
ans.push_back(arr[i]);
i += 1;
}
}
return ans;
}
// Driver code
int main()
{
vector arr({ 2, 2, 2, 100, 5, 4,
2, 9, 10, 88, 24 });
vector ans = merge(arr);
cout << "[";
for(int i= 0; i < ans.size(); i++)
{
if(i == ans.size() - 1)
cout << ans[i] << "]";
else
cout << ans[i] << ", ";
}
}
// This code is contributed by Samarth Java
// Java program to merge the array
// as per the given condition
import java.util.*;
class GFG{
// Function to merge the array
// as per the given condition
static Vector merge(int []arr)
{
// Variable to store the final
// sequence
Vector ans = new Vector();
// Temporary array to store the
// even numbers while traversing
Vector e = new Vector();
int i = 0;
int j = 0;
int count = 0;
// Iterating through the array
while (i < arr.length)
{
// If the element is even
if (arr[i] % 2 == 0)
{
j = i;
// Iterating till an odd element
// is found
while (j < arr.length)
{
// Keep appending into the
// temporary array if the
// even number is occurred
if (arr[j] % 2 == 0)
{
e.add(arr[j]);
count += 1;
}
// Break if an odd number
// has occurred
else
break;
j += 1;
}
// If the series has at least
// three elements, then merge
if (count >= 3)
ans.add(e.get(0));
// Else, add all elements to the
// answer array
else
{
for(int ii : e)
ans.add(ii);
}
// Reseting the count and
// temp array
count = 0;
e.clear();
i = j;
}
// If the element is odd, add
// it to the answer array
else
{
ans.add(arr[i]);
i += 1;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int []arr = { 2, 2, 2, 100, 5, 4,
2, 9, 10, 88, 24 };
Vector ans = merge(arr);
System.out.print("[");
for(int i= 0; i < ans.size(); i++)
{
if (i == ans.size() - 1)
System.out.print(ans.get(i) + "]");
else
System.out.print(ans.get(i) + ", ");
}
}
}
// This code is contributed by Amit Katiyar Python3
# Python3 program to merge the array
# as per the given condition
# Function to merge the array
# as per the given condition
def merge(arr):
# Variable to store the final
# sequence
ans = []
# Temporary array to store the
# even numbers while traversing
e =[]
i = 0
j = 0
count = 0
# Iterating through the array
while i= 3):
ans.append(e[0])
# Else, add all elements to the
# answer array
else:
for i in e:
ans.append(i)
# Reseting the count and
# temp array
count = 0
e =[]
i = j
# If the element is odd, add
# it to the answer array
else:
ans.append(arr[i])
i+= 1
return ans
# Driver code
if __name__ == "__main__":
arr = [2, 2, 2, 100, 5, 4, 2, 9, 10, 88, 24]
print(merge(arr)) C#
// C# program to merge the array
// as per the given condition
using System;
using System.Collections.Generic;
class GFG{
// Function to merge the array
// as per the given condition
static List merge(int []arr)
{
// Variable to store the final
// sequence
List ans = new List();
// Temporary array to store the
// even numbers while traversing
List e = new List();
int i = 0;
int j = 0;
int count = 0;
// Iterating through the array
while (i < arr.Length)
{
// If the element is even
if (arr[i] % 2 == 0)
{
j = i;
// Iterating till an odd element
// is found
while (j < arr.Length)
{
// Keep appending into the
// temporary array if the
// even number is occurred
if (arr[j] % 2 == 0)
{
e.Add(arr[j]);
count += 1;
}
// Break if an odd number
// has occurred
else
break;
j += 1;
}
// If the series has at least
// three elements, then merge
if (count >= 3)
ans.Add(e[0]);
// Else, add all elements to the
// answer array
else
{
foreach(int ii in e)
ans.Add(ii);
}
// Reseting the count and
// temp array
count = 0;
e.Clear();
i = j;
}
// If the element is odd, add
// it to the answer array
else
{
ans.Add(arr[i]);
i += 1;
}
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {2, 2, 2, 100, 5, 4,
2, 9, 10, 88, 24};
List ans = merge(arr);
Console.Write("[");
for(int i= 0; i < ans.Count; i++)
{
if (i == ans.Count - 1)
Console.Write(ans[i] + "]");
else
Console.Write(ans[i] + ", ");
}
}
}
// This code is contributed by 29AjayKumar 输出：
[2, 5, 4, 2, 9, 10]
时间复杂度： O(N 2 ) ，其中N是数组的长度。