打印给定整数数组的所有不同元素
原文： https://www.geeksforgeeks.org/print-distinct-elements-given-integer-array/
方法：
-
将所有输入整数放入哈希映射的键
-
在循环外部打印键集
Java
import java.util.HashMap;
public class UniqueInArray2 {
public static void main(String args[])
{
int ar[] = { 10, 5, 3, 4, 3, 5, 6 };
HashMap<Integer,Integer> hm = new HashMap<Integer,Integer>();
for (int i = 0; i < ar.length; i++) {
hm.put(ar[i], i);
}
// Using hm.keySet() to print output reduces time complexity. - Lokesh
System.out.println(hm.keySet());
}
}
C
// C# implementation of the approach
using System;
using System.Collections.Generic;
public class UniqueInArray2
{
public static void Main(String []args)
{
int []ar = { 10, 5, 3, 4, 3, 5, 6 };
Dictionary<int,int> hm = new Dictionary<int,int>();
for (int i = 0; i < ar.Length; i++)
{
if(hm.ContainsKey(ar[i]))
hm.Remove(ar[i]);
hm.Add(ar[i], i);
}
// Using hm.Keys to print output
// reduces time complexity. - Lokesh
var v = hm.Keys;
foreach(int a in v)
Console.Write(a+" ");
}
}
/* This code contributed by PrinciRaj1992 */
给定一个整数数组，打印数组中所有不同的元素。 给定的数组可能包含重复项，并且输出应仅将每个元素打印一次。 给定的数组未排序。
示例：
Input: arr[] = {12, 10, 9, 45, 2, 10, 10, 45}
Output: 12, 10, 9, 45, 2
Input: arr[] = {1, 2, 3, 4, 5}
Output: 1, 2, 3, 4, 5
Input: arr[] = {1, 1, 1, 1, 1}
Output: 1
一个简单的解决方案是使用两个嵌套循环。 外循环从最左边的元素开始一个接一个地选择一个元素。 内部循环检查元素的左侧是否存在。 如果存在，则忽略该元素，否则打印该元素。 以下是简单算法的实现。
C++
// C++ program to print all distinct elements in a given array
#include <bits/stdc++.h>
using namespace std;
void printDistinct(int arr[], int n)
{
// Pick all elements one by one
for (int i=0; i<n; i++)
{
// Check if the picked element is already printed
int j;
for (j=0; j<i; j++)
if (arr[i] == arr[j])
break;
// If not printed earlier, then print it
if (i == j)
cout << arr[i] << " ";
}
}
// Driver program to test above function
int main()
{
int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
int n = sizeof(arr)/sizeof(arr[0]);
printDistinct(arr, n);
return 0;
}
Java
// Java program to print all distinct
// elements in a given array
import java.io.*;
class GFG {
static void printDistinct(int arr[], int n)
{
// Pick all elements one by one
for (int i = 0; i < n; i++)
{
// Check if the picked element
// is already printed
int j;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;
// If not printed earlier,
// then print it
if (i == j)
System.out.print( arr[i] + " ");
}
}
// Driver program
public static void main (String[] args)
{
int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
int n = arr.length;
printDistinct(arr, n);
}
}
// This code is contributed by vt_m
Python3
# python program to print all distinct
# elements in a given array
def printDistinct(arr, n):
# Pick all elements one by one
for i in range(0, n):
# Check if the picked element
# is already printed
d = 0
for j in range(0, i):
if (arr[i] == arr[j]):
d = 1
break
# If not printed earlier,
# then print it
if (d == 0):
print(arr[i])
# Driver program to test above function
arr = [6, 10, 5, 4, 9, 120, 4, 6, 10]
n = len(arr)
printDistinct(arr, n)
# This code is contributed by Sam007.
C
// C# program to print all distinct
// elements in a given array
using System;
class GFG {
static void printDistinct(int []arr, int n)
{
// Pick all elements one by one
for (int i = 0; i < n; i++)
{
// Check if the picked element
// is already printed
int j;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;
// If not printed earlier,
// then print it
if (i == j)
Console.Write(arr[i] + " ");
}
}
// Driver program
public static void Main ()
{
int []arr = {6, 10, 5, 4, 9, 120,
4, 6, 10};
int n = arr.Length;
printDistinct(arr, n);
}
}
// This code is contributed by Sam007.
PHP
<?php
// PHP program to print all distinct
// elements in a given array
function printDistinct($arr, $n)
{
// Pick all elements one by one
for($i = 0; $i < $n; $i++)
{
// Check if the picked element
// is already printed
$j;
for($j = 0; $j < $i; $j++)
if ($arr[$i] == $arr[$j])
break;
// If not printed
// earlier, then print it
if ($i == $j)
echo $arr[$i] , " ";
}
}
// Driver Code
$arr = array(6, 10, 5, 4, 9, 120, 4, 6, 10);
$n = sizeof($arr);
printDistinct($arr, $n);
// This code is contributed by nitin mittal
?>
输出：
6 10 5 4 9 120
上述解决方案的时间复杂度为O(n ^ 2)。 我们可以使用排序来解决O(nLogn)时间中的问题。 这个想法很简单，首先对数组进行排序，以便每个元素的所有出现变为连续。 一旦出现连续，我们就可以遍历排序后的数组并在O(n)时间内打印出不同的元素。 以下是该想法的实现。
C++
// C++ program to print all distinct elements in a given array
#include <bits/stdc++.h>
using namespace std;
void printDistinct(int arr[], int n)
{
// First sort the array so that all occurrences become consecutive
sort(arr, arr + n);
// Traverse the sorted array
for (int i=0; i<n; i++)
{
// Move the index ahead while there are duplicates
while (i < n-1 && arr[i] == arr[i+1])
i++;
// print last occurrence of the current element
cout << arr[i] << " ";
}
}
// Driver program to test above function
int main()
{
int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
int n = sizeof(arr)/sizeof(arr[0]);
printDistinct(arr, n);
return 0;
}
Java
// Java program to print all distinct
// elements in a given array
import java.io.*;
import java .util.*;
class GFG
{
static void printDistinct(int arr[], int n)
{
// First sort the array so that
// all occurrences become consecutive
Arrays.sort(arr);
// Traverse the sorted array
for (int i = 0; i < n; i++)
{
// Move the index ahead while
// there are duplicates
while (i < n - 1 && arr[i] == arr[i + 1])
i++;
// print last occurrence of
// the current element
System.out.print(arr[i] +" ");
}
}
// Driver program
public static void main (String[] args)
{
int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
int n = arr.length;
printDistinct(arr, n);
}
}
// This code is contributed by vt_m
Python3
# Python program to print all distinct
# elements in a given array
def printDistinct(arr, n):
# First sort the array so that
# all occurrences become consecutive
arr.sort();
# Traverse the sorted array
for i in range(n):
# Move the index ahead while there are duplicates
if(i < n-1 and arr[i] == arr[i+1]):
while (i < n-1 and (arr[i] == arr[i+1])):
i+=1;
# prlast occurrence of the current element
else:
print(arr[i], end=" ");
# Driver code
arr = [6, 10, 5, 4, 9, 120, 4, 6, 10];
n = len(arr);
printDistinct(arr, n);
# This code has been contributed by 29AjayKumar
C
// C# program to print all distinct
// elements in a given array
using System;
class GFG {
static void printDistinct(int []arr, int n)
{
// First sort the array so that
// all occurrences become consecutive
Array.Sort(arr);
// Traverse the sorted array
for (int i = 0; i < n; i++)
{
// Move the index ahead while
// there are duplicates
while (i < n - 1 && arr[i] == arr[i + 1])
i++;
// print last occurrence of
// the current element
Console.Write(arr[i] + " ");
}
}
// Driver program
public static void Main ()
{
int []arr = {6, 10, 5, 4, 9, 120, 4, 6, 10};
int n = arr.Length;
printDistinct(arr, n);
}
}
// This code is contributed by Sam007.
PHP
<?php
// PHP program to print all distinct
// elements in a given array
function printDistinct( $arr, $n)
{
// First sort the array so
// that all occurrences
// become consecutive
sort($arr);
// Traverse the sorted array
for ($i = 0; $i < $n; $i++)
{
// Move the index ahead
// while there are duplicates
while ($i < $n - 1 and
$arr[$i] == $arr[$i + 1])
$i++;
// print last occurrence
// of the current element
echo $arr[$i] , " ";
}
}
// Driver Code
$arr = array(6, 10, 5, 4, 9, 120, 4, 6, 10);
$n = count($arr);
printDistinct($arr, $n);
// This code is contributed by anuj_67.
?>
输出：
4 5 6 9 10 120
我们可以使用散列来平均解决O(n)时间。 这个想法是从左到右遍历给定的数组，并在哈希表中跟踪访问的元素。 以下是该想法的实现。
C++
/* CPP program to print all distinct elements
of a given array */
#include<bits/stdc++.h>
using namespace std;
// This function prints all distinct elements
void printDistinct(int arr[],int n)
{
// Creates an empty hashset
unordered_set<int> s;
// Traverse the input array
for (int i=0; i<n; i++)
{
// If not present, then put it in
// hashtable and print it
if (s.find(arr[i])==s.end())
{
s.insert(arr[i]);
cout << arr[i] << " ";
}
}
}
// Driver method to test above method
int main ()
{
int arr[] = {10, 5, 3, 4, 3, 5, 6};
int n=7;
printDistinct(arr,n);
return 0;
}
Java
/* Java program to print all distinct elements of a given array */
import java.util.*;
class Main
{
// This function prints all distinct elements
static void printDistinct(int arr[])
{
// Creates an empty hashset
HashSet<Integer> set = new HashSet<>();
// Traverse the input array
for (int i=0; i<arr.length; i++)
{
// If not present, then put it in hashtable and print it
if (!set.contains(arr[i]))
{
set.add(arr[i]);
System.out.print(arr[i] + " ");
}
}
}
// Driver method to test above method
public static void main (String[] args)
{
int arr[] = {10, 5, 3, 4, 3, 5, 6};
printDistinct(arr);
}
}
Python3
# Python3 program to print all distinct elements
# of a given array
# This function prints all distinct elements
def printDistinct(arr, n):
# Creates an empty hashset
s = dict();
# Traverse the input array
for i in range(n):
# If not present, then put it in
# hashtable and print it
if (arr[i] not in s.keys()):
s[arr[i]] = arr[i];
print(arr[i], end = " ");
# Driver Code
arr = [10, 5, 3, 4, 3, 5, 6];
n = 7;
printDistinct(arr, n);
# This code is contributed by Princi Singh
C
// C# program to print all distinct
// elements of a given array
using System;
using System.Collections.Generic;
class GFG
{
// This function prints all
// distinct elements
public static void printDistinct(int[] arr)
{
// Creates an empty hashset
HashSet<int> set = new HashSet<int>();
// Traverse the input array
for (int i = 0; i < arr.Length; i++)
{
// If not present, then put it
// in hashtable and print it
if (!set.Contains(arr[i]))
{
set.Add(arr[i]);
Console.Write(arr[i] + " ");
}
}
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = new int[] {10, 5, 3, 4, 3, 5, 6};
printDistinct(arr);
}
}
// This code is contributed by Shrikant13
输出：
10 5 3 4 6
哈希优于排序的另一个优点是，元素的打印顺序与输入数组中的顺序相同。
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