给定一个 m 个元素的数组,我们需要从数组中找到 n 个最小的元素,但它们的顺序必须与给定数组中的顺序相同。
例子:
Input : arr[] = {4, 2, 6, 1, 5},
n = 3
Output : 4 2 1
Explanation :
1, 2 and 4 are 3 smallest numbers and
4 2 1 is their order in given array.
Input : arr[] = {4, 12, 16, 21, 25},
n = 3
Output : 4 12 16
Explanation :
4, 12 and 16 are 3 smallest numbers and
4 12 16 is their order in given array.
制作原始数组的副本,然后对副本数组进行排序。对复制数组进行排序后,保存所有 n 个最小的数字。进一步对于原始数组中的每个元素,如果它出现在 n-smallest 数组中,检查它是否在 n-smallest number 中,然后打印它,否则向前移动。
Make copy_arr[]sort(copy_arr)For all elements in arr[] – Find arr[i] in n-smallest element of copy_arr If found then print the element
以下是上述方法的实现:
C++
// CPP for printing smallest n number in order
#include
using namespace std;
// Function to print smallest n numbers
void printSmall(int arr[], int asize, int n)
{
// Make copy of array
vector copy_arr(arr, arr + asize);
// Sort copy array
sort(copy_arr.begin(), copy_arr.begin() + asize);
// For each arr[i] find whether
// it is a part of n-smallest
// with binary search
for (int i = 0; i < asize; ++i)
if (binary_search(copy_arr.begin(),
copy_arr.begin() + n, arr[i]))
cout << arr[i] << " ";
}
// Driver program
int main()
{
int arr[] = { 1, 5, 8, 9, 6, 7, 3, 4, 2, 0 };
int asize = sizeof(arr) / sizeof(arr[0]);
int n = 5;
printSmall(arr, asize, n);
return 0;
}
Java
// Java for printing smallest n number in order
import java.util.*;
class GFG
{
// Function to print smallest n numbers
static void printSmall(int arr[], int asize, int n)
{
// Make copy of array
int []copy_arr = Arrays.copyOf(arr,asize);
// Sort copy array
Arrays.sort(copy_arr);
// For each arr[i] find whether
// it is a part of n-smallest
// with binary search
for (int i = 0; i < asize; ++i)
{
if (Arrays.binarySearch(copy_arr,0,n, arr[i])>-1)
System.out.print(arr[i] + " ");
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 5, 8, 9, 6, 7, 3, 4, 2, 0 };
int asize = arr.length;
int n = 5;
printSmall(arr, asize, n);
}
}
// This code is contributed by Princi Singh
Python3
# Python3 for printing smallest n number in order
# Function for binary_search
def binary_search(arr, low, high, ele):
while low < high:
mid = (low + high) // 2
if arr[mid] == ele:
return mid
elif arr[mid] > ele:
high = mid
else:
low = mid + 1
return -1
# Function to print smallest n numbers
def printSmall(arr, asize, n):
# Make copy of array
copy_arr = arr.copy()
# Sort copy array
copy_arr.sort()
# For each arr[i] find whether
# it is a part of n-smallest
# with binary search
for i in range(asize):
if binary_search(copy_arr, low = 0,
high = n, ele = arr[i]) > -1:
print(arr[i], end = " ")
# Driver Code
if __name__ == "__main__":
arr = [1, 5, 8, 9, 6, 7, 3, 4, 2, 0]
asize = len(arr)
n = 5
printSmall(arr, asize, n)
# This code is conributed by
# sanjeev2552
C#
// C# for printing smallest n number in order
using System;
class GFG
{
// Function to print smallest n numbers
static void printSmall(int []arr, int asize, int n)
{
// Make copy of array
int []copy_arr = new int[asize];
Array.Copy(arr, copy_arr, asize);
// Sort copy array
Array.Sort(copy_arr);
// For each arr[i] find whether
// it is a part of n-smallest
// with binary search
for (int i = 0; i < asize; ++i)
{
if (Array.BinarySearch(copy_arr, 0, n, arr[i])>-1)
Console.Write(arr[i] + " ");
}
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 5, 8, 9, 6, 7, 3, 4, 2, 0 };
int asize = arr.Length;
int n = 5;
printSmall(arr, asize, n);
}
}
// This code has been contributed by 29AjayKumar
Javascript
输出 :
1 3 4 2 0
时间复杂度: O(n * log(n))
辅助空间: O(n)
为了复制数组,我们需要 O(n) 的空间复杂度,然后为了排序,我们需要 O(n log n) 阶的复杂度。此外,对于 arr[] 中的每个元素,我们正在 copy_arr[] 中执行搜索,这将导致线性搜索的 O(n) 但我们可以通过应用二分搜索来改进它,因此我们的整体时间复杂度将为O(n log n) .
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