打印给定数组中的所有不同字符串

给定一个大小为N的字符串数组arr[] ，任务是打印给定数组中存在的所有不同字符串。

例子：

Input: arr[] = { “Geeks”, “For”, “Geeks”, “Code”, “Coder” } 
Output: Coder Code Geeks For 
Explanation: Since all the strings in the array are distinct, the required output is Coder Code Geeks For.

Input: arr[] = { “Good”, “God”, “Good”, “God”, “god” } 
Output: god Good God

朴素的方法：解决这个问题最简单的方法是根据字符串的字典顺序对数组进行排序。遍历数组并检查数组的当前字符串是否等于先前遍历的字符串。如果发现为假，则打印当前字符串。

时间复杂度： O(N * M * log(N))，其中 M 是最长字符串的长度。
辅助空间： O(1)

高效的方法：为了优化上述方法，想法是使用Hashing。请按照以下步骤解决问题：

• 初始化一个 Set，比如DistString ，以存储来自给定数组的不同字符串。
• 遍历数组并将数组元素插入到DistString 中。
• 最后，从DistString打印所有字符串。

下面是上述方法的实现。

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find the distinct strings
// from the given array
void findDisStr(vector& arr, int N)
{
    // Stores distinct strings
    // from the given array
    unordered_set DistString;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If current string not
        // present into the set
        if (!DistString.count(arr[i])) {
 
            // Insert current string
            // into the set
            DistString.insert(arr[i]);
        }
    }
 
    // Traverse the set DistString
    for (auto String : DistString) {
 
        // Print distinct string
        cout << String << " ";
    }
}
 
// Driver Code
int main()
{
    vector arr = { "Geeks", "For", "Geeks",
                           "Code", "Coder" };
 
    // Stores length of the array
    int N = arr.size();
 
    findDisStr(arr, N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to find the distinct strings
// from the given array
static void findDisStr(List arr, int N)
{
     
    // Stores distinct strings
    // from the given array
    Set DistString = new HashSet();
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // If current string not
        // present into the set
        if (!DistString.contains(arr.get(i)))
        {
             
            // Insert current string
            // into the set
            DistString.add(arr.get(i));
        }
    }
 
    // Traverse the set DistString
    for(String string : DistString)
    {
         
        // Print distinct string
        System.out.print(string + " ");
    }
}
 
// Driver code
public static void main(String[] args)
{
    List arr = Arrays.asList(new String[]{
        "Geeks", "For", "Geeks", "Code", "Coder" });
 
    // Stores length of the array
    int N = arr.size();
 
    findDisStr(arr, N);
}
}
 
// This code is contributed by jithin

Python3

# Python3 program to implement
# the above approach
 
# Function to find the distinct
# strings from the given array
def findDisStr(arr, N):
     
    # Stores distinct strings
    # from the given array
    DistString = set()
 
    # Traverse the array
    for i in range(N):
 
        # If current string not
        # present into the set
        if (arr[i] not in DistString):
 
            # Insert current string
            # into the set
            DistString.add(arr[i])
 
    # Traverse the set DistString
    for string in DistString:
         
        # Print distinct string
        print(string, end = " ")
    
# Driver Code
if __name__ == "__main__":
   
    arr = [ "Geeks", "For", "Geeks",
            "Code", "Coder" ]
 
    # Stores length of the array
    N = len(arr)
 
    findDisStr(arr, N)
     
# This code is contributed by chitranayal

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the distinct strings
// from the given array
static void findDisStr(List arr, int N)
{
     
    // Stores distinct strings
    // from the given array
    HashSet DistString = new HashSet();
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // If current string not
        // present into the set
        if (!DistString.Contains(arr[i]))
        {
             
            // Insert current string
            // into the set
            DistString.Add(arr[i]);
        }
    }
 
    // Traverse the set DistString
    foreach(string a in DistString)
    { 
      Console.Write(a +" "); 
    } 
}
 
// Driver code
public static void Main(String[] args)
{
    List arr = new List(new []{
        "Geeks", "For", "Geeks", "Code", "Coder" });
 
    // Stores length of the array
    int N = arr.Count;
 
    findDisStr(arr, N);
}
}
 
// This code is contributed by jana_sayantan

Javascript

 
// JavaScript program to implement
// the above approach
 
 
// Function to find the distinct strings
// from the given array
function findDisStr(arr, N) {
    // Stores distinct strings
    // from the given array
    let DistString = new Set();
 
    // Traverse the array
    for (let i = N - 1; i >= 0; i--) {
        // If current string not
        // present into the set
        if (!DistString.has(arr[i])) {
 
            // Insert current string
            // into the set
            DistString.add(arr[i]);
        }
    }
     
 
    for (let String of DistString) {
 
        // Print distinct string
        document.write(String + " ");
    }
}
 
// Driver Code
 
let arr = ["Geeks", "For", "Geeks",
    "Code", "Coder"];
 
// Stores length of the array
let N = arr.length;
 
findDisStr(arr, N);
 

输出：

Coder Code Geeks For

时间复杂度： O(N * M)，其中 M 是最长字符串的长度。
辅助空间： O(N * M)

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