给定由N个不同的正整数组成的数组arr [] ，任务是从给定数组中重复查找最小峰元素，然后删除该元素，直到删除所有数组元素。
Peak Element: Any element in the array is know as the peak element based on the following conditions:
- If arr[i – 1] < arr[i] > arr[i + 1], where 1 < i < N – 1, then arr[i] is the peak element.
- If arr[0] > arr[1], then arr[0] is the peak element, where N is the size of the array.
- If arr[N – 2] < arr[N – 1], then arr[N – 1] is the peak element, where N is the size of the array.
If more than one peak element exists in the array, then the minimum value among them needs to be printed.
例子：
Input: arr[] = {1, 9, 7, 8, 2, 6}
Output: [6, 8, 9, 7, 2, 1]
Explanation:
First min peak = 6, as 2 < 6.
The array after removing min peak will be [1, 9, 7, 8, 2].
Second min peak = 8, as 7 < 8 > 2.
The array after removing min peak will be [1, 9, 7, 2]
Third min peak = 9, as 1 < 9 > 7.
The array after removing min peak will be [1, 7, 2]
Fourth min peak = 7, as 1 < 7 > 2.
The array after removing min peak will be [1, 2]
Fifth min peak = 2, as 1 < 2.
The array after removing min peak will be [1]
Sixth min peak = 1.
Therefore, the list of minimum peak is [6, 8, 9, 7, 2, 1].
Input: arr []= {1, 5, 3, 7, 2}
Output: [5, 7, 3, 2, 1]
Explanation:
First min peak = 5, as 1 < 5 > 3.
The array after removing min peak will be [1, 3, 7, 2]
Second min peak = 7, as 3 < 7 > 2.
The array after removing min peak will be [1, 3, 2]
Third min peak = 3, as 1 < 3 > 2.
The array after removing min peak will be [1, 2]
Fourth min peak = 2, as 1 < 2.
The array after removing min peak will be [1]
Fifth min peak = 1.
Therefore, the list of minimum peak is [5, 7, 3, 2, 1]
方法：这个想法是通过使用两个嵌套循环在数组上进行迭代来找到数组的最小峰值元素，其中，外循环指向当前元素，而内循环执行以找到最小峰值元素的索引，删除该峰值数组中的元素并将当前峰值元素存储在结果列表中。完成上述步骤后，打印列表中存储的所有最小峰元素。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to return the list of
// minimum peak elements
void minPeaks(vectorlist)
{
// Length of original list
int n = list.size();
// Initialize resultant list
vectorresult;
// Traverse each element of list
for(int i = 0; i < n; i++)
{
int min = INT_MAX;
int index = -1;
// Length of original list after
// removing the peak element
int size = list.size();
// Traverse new list after removal
// of previous min peak element
for(int j = 0; j < size; j++)
{
// Update min and index,
// if first element of
// list > next element
if (j == 0 && j + 1 < size)
{
if (list[j] > list[j + 1] &&
min > list[j])
{
min = list[j];
index = j;
}
}
else if (j == size - 1 &&
j - 1 >= 0)
{
// Update min and index,
// if last elemnt of
// list > previous one
if (list[j] > list[j - 1] &&
min > list[j])
{
min = list[j];
index = j;
}
}
// Update min and index, if
// list has single element
else if (size == 1)
{
min = list[j];
index = j;
}
// Update min and index,
// if current element >
// adjacent elements
else if (list[j] > list[j - 1] &&
list[j] > list[j + 1] &&
min > list[j])
{
min = list[j];
index = j;
}
}
// Remove current min peak
// element from list
list.erase(list.begin() + index);
// Insert min peak into
// resultant list
result.push_back(min);
}
// Print resultant list
cout << "[";
for(int i = 0; i < result.size(); i++)
{
cout << result[i] << ", ";
}
cout << "]";
}
// Driver Code
int main()
{
// Given array arr[]
vector arr = { 1, 9, 7, 8, 2, 6 };
// Function call
minPeaks(arr);
}
// This code is contributed by bikram2001jha Java
// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG {
// Function to return the list of
// minimum peak elements
static void
minPeaks(ArrayList list)
{
// Length of original list
int n = list.size();
// Initialize resultant list
ArrayList result
= new ArrayList<>();
// Traverse each element of list
for (int i = 0; i < n; i++) {
int min = Integer.MAX_VALUE;
int index = -1;
// Length of original list after
// removing the peak element
int size = list.size();
// Traverse new list after removal
// of previous min peak element
for (int j = 0; j < size; j++) {
// Update min and index,
// if first element of
// list > next element
if (j == 0 && j + 1 < size) {
if (list.get(j) > list.get(j + 1)
&& min > list.get(j)) {
min = list.get(j);
index = j;
}
}
else if (j == size - 1
&& j - 1 >= 0) {
// Update min and index,
// if last elemnt of
// list > previous one
if (list.get(j)
> list.get(j - 1)
&& min
> list.get(j)) {
min = list.get(j);
index = j;
}
}
// Update min and index, if
// list has single element
else if (size == 1) {
min = list.get(j);
index = j;
}
// Update min and index,
// if current element >
// adjacent elements
else if (list.get(j)
> list.get(j - 1)
&& list.get(j)
> list.get(j + 1)
&& min
> list.get(j)) {
min = list.get(j);
index = j;
}
}
// Remove current min peak
// element from list
list.remove(index);
// Insert min peak into
// resultant list
result.add(min);
}
// Print resultant list
System.out.println(result);
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
ArrayList arr = new ArrayList<>(
Arrays.asList(1, 9, 7, 8, 2, 6));
// Function Call
minPeaks(arr);
}
} Python3
# Python3 program for
# the above approach
import sys
# Function to return the list of
# minimum peak elements
def minPeaks(list1):
# Length of original list
n = len(list1)
# Initialize resultant list
result = []
# Traverse each element of list
for i in range (n):
min = sys.maxsize
index = -1
# Length of original list
# after removing the peak
# element
size = len(list1)
# Traverse new list after removal
# of previous min peak element
for j in range (size):
# Update min and index,
# if first element of
# list > next element
if (j == 0 and j + 1 < size):
if (list1[j] > list1[j + 1] and
min > list1[j]):
min = list1[j];
index = j;
elif (j == size - 1 and
j - 1 >= 0):
# Update min and index,
# if last elemnt of
# list > previous one
if (list1[j] > list1[j - 1] and
min > list1[j]):
min = list1[j]
index = j
# Update min and index, if
# list has single element
elif (size == 1):
min = list1[j]
index = j
# Update min and index,
# if current element >
# adjacent elements
elif (list1[j] > list1[j - 1] and
list1[j] > list1[j + 1] and
min > list1[j]):
min = list1[j]
index = j
# Remove current min peak
# element from list
list1.pop(index)
# Insert min peak into
# resultant list
result.append(min)
# Print resultant list
print (result)
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [1, 9, 7, 8, 2, 6]
# Function call
minPeaks(arr)
# This code is contributed by ChitranayalC#
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to return the list of
// minimum peak elements
static void minPeaks(List list)
{
// Length of original list
int n = list.Count;
// Initialize resultant list
List result = new List();
// Traverse each element of list
for (int i = 0; i < n; i++)
{
int min = int.MaxValue;
int index = -1;
// Length of original list after
// removing the peak element
int size = list.Count;
// Traverse new list after removal
// of previous min peak element
for (int j = 0; j < size; j++)
{
// Update min and index,
// if first element of
// list > next element
if (j == 0 && j + 1 < size)
{
if (list[j] > list[j + 1] &&
min > list[j])
{
min = list[j];
index = j;
}
}
else if (j == size - 1 && j - 1 >= 0)
{
// Update min and index,
// if last elemnt of
// list > previous one
if (list[j] > list[j - 1] &&
min > list[j])
{
min = list[j];
index = j;
}
}
// Update min and index, if
// list has single element
else if (size == 1)
{
min = list[j];
index = j;
}
// Update min and index,
// if current element >
// adjacent elements
else if (list[j] > list[j - 1] &&
list[j] > list[j + 1] &&
min > list[j])
{
min = list[j];
index = j;
}
}
// Remove current min peak
// element from list
list.RemoveAt(index);
// Insert min peak into
// resultant list
result.Add(min);
}
// Print resultant list
for (int i = 0; i < result.Count; i++)
Console.Write(result[i] +", ");
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
List arr = new List{1, 9, 7,
8, 2, 6};
// Function Call
minPeaks(arr);
}
}
// This code is contributed by 29AjayKumar 输出：
[6, 8, 9, 7, 2, 1]
时间复杂度： O(N 2 )
辅助空间： O(N)