大小为 K 的最大连续子阵列
原文:https://www . geesforgeks . org/maximum-continuous-sub-array-of-size-k/
给定一个整数数组 arr[] 和一个整数 K ，任务是找到大小最大的连续子数组 K 。如果两个子阵列中的第一个非匹配元素在 X 中的值大于在 Y 中的值，则称子阵列 X 大于子阵列 Y 。
示例:
输入: arr[] = {1，4，3，2，5}，K = 4 输出: 4 3 2 5 两个子阵分别为{1，4，3，2}和{4，3，2，5}。 因此，较大的一个是{4，3，2，5}
输入: arr[] = {1，9，2，7，9，3}，K = 3 T3】输出: 9 2 7
方法:生成所有大小为 K 的子数组，并将它们存储在任何数据结构中。对所有子数组进行排序，答案将是排序后的数据结构中的最后一个子数组。在 C++中，我们可以使用向量的向量来存储大小为 K 的子数组。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns the sub-array
vector<int> findSubarray(int a[], int k, int n)
{
// Data-structure to store all
// the sub-arrays of size K
vector<vector<int> > vec;
// Iterate to find all the sub-arrays
for (int i = 0; i < n - k + 1; i++) {
vector<int> temp;
// Store the sub-array elements in the array
for (int j = i; j < i + k; j++) {
temp.push_back(a[j]);
}
// Push the vector in the container
vec.push_back(temp);
}
// Sort the vector of elements
sort(vec.begin(), vec.end());
// The last sub-array in the sorted order
// will be the answer
return vec[vec.size() - 1];
}
// Driver code
int main()
{
int a[] = { 1, 4, 3, 2, 5 };
int k = 4;
int n = sizeof(a) / sizeof(a[0]);
// Get the sub-array
vector<int> ans = findSubarray(a, k, n);
for (auto it : ans)
cout << it << " ";
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG{
// Function that returns the sub-array
static ArrayList<Integer> findSubarray(int a[],
int k, int n)
{
// Data-structure to store all
// the sub-arrays of size K
ArrayList<
ArrayList<Integer>> vec = new ArrayList<
ArrayList<Integer>>();
// Iterate to find all the sub-arrays
for(int i = 0; i < n - k + 1; i++)
{
ArrayList<Integer> temp = new ArrayList<Integer>();
// Store the sub-array elements in the array
for(int j = i; j < i + k; j++)
{
temp.add(a[j]);
}
// Push the vector in the container
vec.add(temp);
}
// Sort the vector of elements
Collections.sort(vec, new Comparator<ArrayList<Integer>>()
{
@Override
public int compare(ArrayList<Integer> o1,
ArrayList<Integer> o2)
{
return o1.get(0).compareTo(o2.get(0));
}
});
// The last sub-array in the sorted order
// will be the answer
return vec.get(vec.size() - 1);
}
// Driver code
public static void main(String[] args)
{
int a[] = { 1, 4, 3, 2, 5 };
int k = 4;
int n = a.length;
// Get the sub-array
ArrayList<Integer> ans = findSubarray(a, k, n);
for(int it: ans)
{
System.out.print(it + " ");
}
}
}
// This code is contributed by avanitrachhadiya2155
Python 3
# Python3 implementation of the approach
# Function that returns the sub-array
def findSubarray(a, k, n):
# Data-structure to store all
# the sub-arrays of size K
vec=[]
# Iterate to find all the sub-arrays
for i in range(n-k+1):
temp=[]
# Store the sub-array elements in the array
for j in range(i,i+k):
temp.append(a[j])
# Push the vector in the container
vec.append(temp)
# Sort the vector of elements
vec=sorted(vec)
# The last sub-array in the sorted order
# will be the answer
return vec[len(vec) - 1]
# Driver code
a =[ 1, 4, 3, 2, 5 ]
k = 4
n = len(a)
# Get the sub-array
ans = findSubarray(a, k, n)
for it in ans:
print(it,end=" ")
# This code is contributed by mohit kumar 29
C
// C# implementation of the approach
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG
{
// Function that returns the sub-array
static List<int> findSubarray(int[] a, int k, int n)
{
// Data-structure to store all
// the sub-arrays of size K
List<List<int>> vec = new List<List<int>>();
// Iterate to find all the sub-arrays
for(int i = 0; i < n - k + 1; i++)
{
List<int> temp = new List<int>();
// Store the sub-array elements in the array
for(int j = i; j < i + k; j++)
{
temp.Add(a[j]);
}
// Push the vector in the container
vec.Add(temp);
}
// Sort the vector of elements
vec.OrderBy( l => l[0]);
// The last sub-array in the sorted order
// will be the answer
return vec[vec.Count - 1];
}
// Driver code
static public void Main (){
int[] a = { 1, 4, 3, 2, 5 };
int k = 4;
int n = a.Length;
// Get the sub-array
List<int> ans = findSubarray(a, k, n);
foreach(int it in ans)
{
Console.Write(it + " ");
}
}
}
// This code is contributed by rag2127
java 描述语言
<script>
// Javascript implementation of the approach
// Function that returns the sub-array
function findSubarray(a, k, n)
{
// Data-structure to store all
// the sub-arrays of size K
var vec = [];
// Iterate to find all the sub-arrays
for (var i = 0; i < n - k + 1; i++) {
var temp = [];
// Store the sub-array elements in the array
for (var j = i; j < i + k; j++) {
temp.push(a[j]);
}
// Push the vector in the container
vec.push(temp);
}
// Sort the vector of elements
vec.sort()
// The last sub-array in the sorted order
// will be the answer
return vec[vec.length - 1];
}
// Driver code
var a = [1, 4, 3, 2, 5];
var k = 4;
var n = a.length;
// Get the sub-array
var ans = findSubarray(a, k, n);
ans.forEach(it => {
document.write(it+ " ")
});
// This coe is contributed by noob2000.
</script>
Output:
4 3 2 5
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