给定大小为N的数组arr [] ，任务是重新排列数组元素，以使数组中局部最小值的数量最大。
注意：如果一个元素arr [x]小于或等于它的两个相邻元素，则称其为局部最小值。第一个和最后一个数组元素不能视为局部最小值。
例子：
Input: arr[]= {1, 2, 3, 4, 5}
Output: 3 1 4 2 5
Explanation:
Rearranging array elements to {3, 1, 4, 2, 5}. The count of local minima in the array is 2, i.e. {arr[1], arr[3]}, which is the maximum possible count of local minima that can be obtained from the array. Therefore, the required output is 3 1 4 2 5.
Input: arr[]= {1, 2, 3, 4, 5, 6}
Output: 4 1 5 2 6 3
方法：想法是使用排序算法和两个指针技术来解决此问题。请按照以下步骤解决此问题：
- 以升序对数组进行排序。
- 初始化两个变量，例如left = 0和right = N / 2 ，分别存储左指针和右指针的索引。
- 遍历数组，并在每次遍历时，首先打印arr [right]的值，然后打印arr [left]的值，并将left和right的值增加1 。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to rearrange array elements to
// maximize count of local minima in the array
void rearrangeArrMaxcntMinima(int arr[], int N)
{
// Sort the array in
// ascending order
sort(arr, arr + N);
// Stores index of
// left pointer
int left = 0;
// Stores index of
// right pointer
int right = N / 2;
// Traverse the array elements
while (left < N / 2 || right < N) {
// if right is less than N
if (right < N) {
// Print array element
cout << arr[right] << " ";
// Update right
right++;
}
if (left < N / 2) {
// Print array element
cout << arr[left] << " ";
// Update left
left++;
}
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
rearrangeArrMaxcntMinima(arr, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to rearrange array elements to
// maximize count of local minima in the array
static void rearrangeArrMaxcntMinima(int arr[],
int N)
{
// Sort the array in
// ascending order
Arrays.sort(arr);
// Stores index of
// left pointer
int left = 0;
// Stores index of
// right pointer
int right = N / 2;
// Traverse the array elements
while (left < N / 2 || right < N)
{
// If right is less than N
if (right < N)
{
// Print array element
System.out.print(arr[right] + " ");
// Update right
right++;
}
if (left < N / 2)
{
// Print array element
System.out.print(arr[left] + " ");
// Update left
left++;
}
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = arr.length;
rearrangeArrMaxcntMinima(arr, N);
}
}
// This code is contributed by code_huntPython3
# Python3 program to implement
# the above approach
# Function to rearrange array
# elements to maximize count of
# local minima in the array
def rearrangeArrMaxcntMinima(arr, N):
# Sort the array in
# ascending order
arr.sort()
# Stores index of
# left pointer
left = 0
# Stores index of
# right pointer
right = N // 2
# Traverse the array elements
while (left < N // 2 or
right < N):
# if right is less
# than N
if (right < N):
# Print array element
print(arr[right],
end = " ")
# Update right
right += 1
if (left < N // 2):
# Print array element
print(arr[left],
end = " ")
# Update left
left += 1
# Driver Code
if __name__ == "__main__":
arr = [1, 2, 3, 4, 5]
N = len(arr)
rearrangeArrMaxcntMinima(arr, N)
# This code is contributed by ChitranayalC#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to rearrange array elements to
// maximize count of local minima in the array
static void rearrangeArrMaxcntMinima(int []arr,
int N)
{
// Sort the array in
// ascending order
Array.Sort(arr);
// Stores index of
// left pointer
int left = 0;
// Stores index of
// right pointer
int right = N / 2;
// Traverse the array elements
while (left < N / 2 || right < N)
{
// If right is less than N
if (right < N)
{
// Print array element
Console.Write(arr[right] + " ");
// Update right
right++;
}
if (left < N / 2)
{
// Print array element
Console.Write(arr[left] + " ");
// Update left
left++;
}
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4, 5 };
int N = arr.Length;
rearrangeArrMaxcntMinima(arr, N);
}
}
// This code is contributed by Rajput-Ji输出：
3 1 4 2 5
时间复杂度： O(N log(N))
辅助空间： O(1)