根据最后出现的顺序从字符串删除重复项

给定一个字符串，从字符串中删除重复的字符，保留了重复的字符最后一次出现。假设字符区分大小写。
例子：

Input : geeksforgeeks 
Output : forgeks 
Explanation : Please note that we keep only last occurrences of repeating characters in same order as they appear in input. If we see result from right side, we can notice that we keep last ‘s’, then last ‘k’ , and so on.
Input : hi this is sample test 
Output : hiampl est 
Explanation : Here, the output contains last occurrence of every character, even ” “(spaces), and removing the duplicates. Like in this example, there are 4 spaces count, so we have only the last occurrence of space in it removing the others. And there is only last occurrence of each character without repetition. 
Input : Abcda 
Output : Abcda

天真的解决方案：从左到右遍历给定的字符串。对于每个字符，检查它是否也出现在右侧。如果是，则不包含在输出中，否则包含它。

C++

// C++ program to remove duplicate character
// from character array and print in sorted
// order
#include 
using namespace std;
 
string removeDuplicates(string str)
{
    // Used as index in the modified string
    int n = str.length();
 
    // Traverse through all characters
    string res = "";
    for (int i = 0; i < n; i++) {
 
        // Check if str[i] is present before it
        int j;
        for (j = i+1; j < n; j++)
            if (str[i] == str[j])
                break;
 
        // If not present, then add it to
        // result.
        if (j == n)
            res = res + str[i];
    }
    return res;
}
 
// Driver code
int main()
{
    string str = "geeksforgeeks";
    cout << removeDuplicates(str);
    return 0;
}

Java

// Java program to remove duplicate character
// from character array and print in sorted
// order
 
import java.util.*;
 
class GFG{
 
static String removeDuplicates(String str)
{
    // Used as index in the modified String
    int n = str.length();
 
    // Traverse through all characters
    String res = "";
    for (int i = 0; i < n; i++)
    {
        // Check if str[i] is present before it
        int j;
        for (j = i + 1; j < n; j++)
            if (str.charAt(i) == str.charAt(j))
                break;
 
        // If not present, then add it to
        // result.
        if (j == n)
            res = res + str.charAt(i);
    }
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    String str = "geeksforgeeks";
    System.out.print(removeDuplicates(str));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to remove duplicate character
# from character array and prin sorted
# order
 
def removeDuplicates(str):
      
    # Used as index in the modified string
    n = len(str)
 
    # Traverse through all characters
    res = ""
    for i in range(n):
 
        # Check if str[i] is present before it
        j = i + 1
        while j < n:
            if (str[i] == str[j]):
                break
            j += 1
 
        # If not present, then add it to
        # result.
        if (j == n):
            res = res + str[i]
    return res
 
# Driver code
if __name__ == '__main__':
    str = "geeksforgeeks"
    print(removeDuplicates(str))
 
# This code is contributed by mohit kumar 29

C#

// C# program to remove duplicate character
// from character array and print in sorted
// order
using System;
 
class GFG{
 
static String removeDuplicates(String str)
{
     
    // Used as index in the modified String
    int n = str.Length;
 
    // Traverse through all characters
    String res = "";
    for(int i = 0; i < n; i++)
    {
 
       // Check if str[i] is present before it
       int j;
       for(j = i + 1; j < n; j++)
          if (str[i] == str[j])
              break;
 
       // If not present, then add it to
       // result.
       if (j == n)
           res = res + str[i];
    }
     
    return res;
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "geeksforgeeks";
    Console.Write(removeDuplicates(str));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// Javascript program to remove duplicate character
// from character array and print in sorted
// order
 
  function removeDuplicates(str)
{
    // Used as index in the modified String
    let n = str.length;
   
    // Traverse through all characters
    let res = "";
    for (let i = 0; i < n; i++)
    {
        // Check if str[i] is present before it
        let j;
        for (j = i + 1; j < n; j++)
            if (str[i] == str[j])
                break;
   
        // If not present, then add it to
        // result.
        if (j == n)
            res = res + str[i];
    }
    return res;
}
 
// Driver code
     
    let str = "geeksforgeeks";
    document.write(removeDuplicates(str));
 
// This code is contributed by code_hunt.

输出：

forgeks

时间复杂度：O(n*n)
有效的解决方案：这个想法是使用散列。
1）初始化一个空的hash表，res =“”
2) 从右到左遍历输入字符串。如果当前字符不在哈希表中，则将其附加到 res 并将其插入哈希表中。否则无视。

C++

// C++ program to remove duplicate character
// from character array and print in sorted
// order
#include 
using namespace std;
 
string removeDuplicates(string str)
{
    // Used as index in the modified string
    int n = str.length();
 
    // Create an empty hash table
    unordered_set s;
 
    // Traverse through all characters from
    // right to left
    string res = "";
    for (int i = n-1; i >= 0; i--) {
 
       // If current character is not in
       if (s.find(str[i]) == s.end())
       {
          res = res + str[i];
          s.insert(str[i]);
       }
    }
 
    // Reverse the result string
    reverse(res.begin(), res.end());
 
    return res;
}
 
// Driver code
int main()
{
    string str = "geeksforgeeks";
    cout << removeDuplicates(str);
    return 0;
}

Java

// Java program to remove duplicate character
// from character array and print in sorted
// order
import java.util.*;
 
class GFG{
 
static String removeDuplicates(String str)
{
     
    // Used as index in the modified String
    int n = str.length();
 
    // Create an empty hash table
    HashSet s = new HashSet();
 
    // Traverse through all characters from
    // right to left
    String res = "";
    for(int i = n - 1; i >= 0; i--)
    {
         
       // If current character is not in
       if (!s.contains(str.charAt(i)))
       {
           res = res + str.charAt(i);
           s.add(str.charAt(i));
       }
    }
     
    // Reverse the result String
    res = reverse(res);
    return res;
}
 
static String reverse(String input)
{
     
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
     
    for(l = 0; l < r; l++, r--)
    {
       char temp = a[l];
       a[l] = a[r];
       a[r] = temp;
    }
    return String.valueOf(a);
}
 
// Driver code
public static void main(String[] args)
{
    String str = "geeksforgeeks";
    System.out.print(removeDuplicates(str));
}
}
 
// This code is contributed by sapnasingh4991

蟒蛇3

# Python3 program to remove duplicate character
# from character array and prin sorted order
def removeDuplicates(str):
     
    # Used as index in the modified string
    n = len(str)
     
    # Create an empty hash table
    s = set()
     
    # Traverse through all characters from
    # right to left
    res = ""
    for i in range(n - 1, -1, -1):
         
        # If current character is not in
        if (str[i] not in s):
            res = res + str[i]
            s.add(str[i])
     
    # Reverse the result string
    res = res[::-1]
    return res
 
# Driver code
str = "geeksforgeeks"
 
print(removeDuplicates(str))
 
# This code is contributed by ShubhamCoder

C＃

// C# program to remove duplicate character
// from character array and print in sorted
// order
using System;
using System.Collections.Generic;
 
class GFG{
 
static String removeDuplicates(String str)
{
     
    // Used as index in the modified String
    int n = str.Length;
 
    // Create an empty hash table
    HashSet s = new HashSet();
 
    // Traverse through all characters
    // from right to left
    String res = "";
    for(int i = n - 1; i >= 0; i--)
    {
         
        // If current character is not in
        if (!s.Contains(str[i]))
        {
            res = res + str[i];
            s.Add(str[i]);
        }
    }
     
    // Reverse the result String
    res = reverse(res);
    return res;
}
 
static String reverse(String input)
{
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
     
    for(l = 0; l < r; l++, r--)
    {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("", a);
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "geeksforgeeks";
    Console.Write(removeDuplicates(str));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program to remove duplicate character
// from character array and print in sorted
// order
 
function removeDuplicates(str)
{
    // Used as index in the modified string
    var n = str.length;
 
    // Create an empty hash table
    var s = new Set();
 
    // Traverse through all characters from
    // right to left
    var res = "";
    for (var i = n-1; i >= 0; i--) {
 
       // If current character is not in
       if (!s.has(str[i]))
       {
          res = res + str[i];
          s.add(str[i]);
       }
    }
 
    // Reverse the result string
    res = res.split('').reverse().join('');
 
    return res;
}
 
// Driver code
var str = "geeksforgeeks";
document.write( removeDuplicates(str));
 
// This code is contributed by rrrtnx.

输出：

forgeks

时间复杂度：O(n)

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