按顺序打印字符串的所有不同字符(3种方法)

给定一个字符串，找到其中所有不同的(或非重复的字符)。例如，如果输入字符串为“ Geeks for Geeks”，则输出应为“ for”，如果输入字符串为“ Geeks Quiz”，则输出应为“ GksQuiz”。

不同的字符应该按照在输入字符串出现的相同顺序打印。

例子：

Input  : Geeks for Geeks
Output : for

Input  : Hello Geeks
Output : HoGks

方法1(简单：O(n ² ))
一个简单的解决方案是运行两个循环。从左侧开始遍历。对于每个字符，请检查是否重复。如果字符不重复，则增加非重复字符的计数。当计数变为1时，返回每个字符。

方法2(有效，但需要两个遍历：O(n))

1. 创建一个数组count []来存储字符。
2. 遍历输入字符串str，并对每个字符x = str [i]执行以下操作。
   增量计数[x]。
3. 再次遍历输入字符串，并对每个字符str [i]执行以下操作
   1. 如果count [x]为1，则输出唯一字符
   2. 如果count [x]大于1，则忽略重复的字符。

以下是上述想法的实现。

C++

// C++ program to print distinct characters of a
// string.
# include 
using namespace std;
# define NO_OF_CHARS 256
  
/* Print duplicates present in the passed string */
void printDistinct(char *str)
{
    // Create an array of size 256 and count of
    // every character in it
    int count[NO_OF_CHARS];
  
    /* Count array with frequency of characters */
    int i;
    for (i = 0; *(str+i); i++)
        if(*(str+i)!=' ')
            count[*(str+i)]++;
    int n = i;
  
    // Print characters having count more than 0
    for (i = 0; i < n; i++)
        if (count[*(str+i)] == 1)
            cout<< str[i];
}
  
/* Driver program*/
int main()
{
    char str[] = "GeeksforGeeks";
    printDistinct((str);
    return 0;
}

Java

// Java program to print distinct characters of a
// string.
public class GFG {
    static final int NO_OF_CHARS = 256;
       
    /* Print duplicates present in the passed string */
    static void printDistinct(String str)
    {
        // Create an array of size 256 and count of
        // every character in it
        int[] count = new int[NO_OF_CHARS];
       
        /* Count array with frequency of characters */
        int i;
        for (i = 0; i < str.length(); i++)
            if(str.charAt(i)!=' ')
                count[(int)str.charAt(i)]++;
        int n = i;
       
        // Print characters having count more than 0
        for (i = 0; i < n; i++)
            if (count[(int)str.charAt(i)] == 1)
                System.out.print(str.charAt(i));
    }
       
    /* Driver program*/
    public static void main(String args[])
    {
        String str = "GeeksforGeeks";
        printDistinct(str);
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Python3 program to print distinct 
# characters of a string.
NO_OF_CHARS = 256
  
# Print duplicates present in the 
# passed string 
def printDistinct(str):
  
    # Create an array of size 256 and 
    # count of every character in it
    count = [0] * NO_OF_CHARS
  
    # Count array with frequency of 
    # characters 
    for i in range (len(str)):
        if(str[i] != ' '):
            count[ord(str[i])] += 1
    n = i
  
    # Print characters having count 
    # more than 0
    for i in range(n):
        if (count[ord(str[i])] == 1):
            print (str[i], end = "")
  
# Driver Code
if __name__ == "__main__":
  
    str = "GeeksforGeeks"
    printDistinct(str)
      
# This code is contributed by ita_c

C#

// C# program to print distinct characters
// of a string.
using System;
  
public class GFG {
      
    static int NO_OF_CHARS = 256;
      
    /* Print duplicates present in the
    passed string */
    static void printDistinct(String str)
    {
          
        // Create an array of size 256 and
        // count of every character in it
        int[] count = new int[NO_OF_CHARS];
      
        /* Count array with frequency of
        characters */
        int i;
          
        for (i = 0; i < str.Length; i++)
            if(str[i]!=' ')
                count[(int)str[i]]++;
                  
        int n = i;
      
        // Print characters having count
        // more than 0
        for (i = 0; i < n; i++)
            if (count[(int)str[i]] == 1)
                Console.Write(str[i]);
    }
      
    /* Driver program*/
    public static void Main()
    {
        String str = "GeeksforGeeks";
          
        printDistinct(str);
    }
}
  
// This code is contributed by parashar.

---

输出：

for

方法3(O(n)，并且需要一个遍历)
这个想法是使用两个大小为256的辅助数组(假设字符使用8位存储)。

1. 将count []中的所有值初始化为0并将index []中的所有值初始化为n，其中n是字符串的长度。
2. 遍历输入字符串str，并对每个字符c = str [i]执行以下操作。
   • 增量计数[x]。
   • 如果count [x]为1，则将x的索引存储在index [x]中，即index [x] = i
   • 如果count [x]为2，则从index []中删除x，即index [x] = n
3. 现在index []具有所有不同字符的索引。排序索引并使用它打印字符。请注意，假设字符数是固定的(通常为256)，此步骤将花费O(1)时间

以下是上述想法的实现。

C++

// C++ program to find all distinct characters
// in a string
#include 
using namespace std;
const int MAX_CHAR = 256;
  
// Function to print distinct characters in
// given string str[]
void printDistinct(string str)
{
    int n = str.length();
  
    // count[x] is going to store count of
    // character 'x' in str. If x is not present,
    // then it is going to store 0.
    int count[MAX_CHAR];
  
    // index[x] is going to store index of character
    // 'x' in str. If x is not present or x is
    // more than once, then it is going to store a value
    // (for example, length of string) that cannot be
    // a valid index in str[]
    int index[MAX_CHAR];
  
    // Initialize counts of all characters and indexes
    // of distinct characters.
    for (int i = 0; i < MAX_CHAR; i++)
    {
        count[i] = 0;
        index[i] = n; // A value more than any index
                      // in str[]
    }
  
    // Traverse the input string
    for (int i = 0; i < n; i++)
    {
        // Find current character and increment its
        // count
        char x = str[i];
        ++count[x];
  
        // If this is first occurrence, then set value
        // in index as index of it.
        if (count[x] == 1 && x !=' ')
            index[x] = i;
  
        // If character repeats, then remove it from
        // index[]
        if (count[x] == 2)
            index[x] = n;
    }
  
    // Since size of index is constant, below operations
    // take constant time.
    sort(index, index+MAX_CHAR);
    for (int i=0; i

Java

// Java program to print distinct characters of 
// a string.
import java.util.Arrays;
  
public class GFG {
      
    static final int MAX_CHAR = 256;
       
    // Function to print distinct characters in
    // given string str[]
    static void printDistinct(String str)
    {
        int n = str.length();
       
        // count[x] is going to store count of
        // character 'x' in str. If x is not present,
        // then it is going to store 0.
        int[] count = new int[MAX_CHAR];
       
        // index[x] is going to store index of character
        // 'x' in str. If x is not present or x is
        // more than once, then it is going to store a 
        // value (for example, length of string) that 
        // cannot be a valid index in str[]
        int[] index = new int[MAX_CHAR];
       
        // Initialize counts of all characters and 
        // indexes of distinct characters.
        for (int i = 0; i < MAX_CHAR; i++)
        {
            count[i] = 0;
            index[i] = n; // A value more than any 
                          // index in str[]
        }
       
        // Traverse the input string
        for (int i = 0; i < n; i++)
        {
            // Find current character and increment 
            // its count
            char x = str.charAt(i);
            ++count[x];
       
            // If this is first occurrence, then set 
            // value in index as index of it.
            if (count[x] == 1 && x !=' ')
                index[x] = i;
       
            // If character repeats, then remove it 
            // from index[]
            if (count[x] == 2)
                index[x] = n;
        }
       
        // Since size of index is constant, below 
        // operations take constant time.
        Arrays.sort(index);
          
        for (int i = 0; i < MAX_CHAR && index[i] != n;
                                                  i++)
           System.out.print(str.charAt(index[i]));
    }
       
    // Driver code
    public static void main(String args[])
    {
        String str = "GeeksforGeeks";
        printDistinct(str);
    }
}
// This code is contributed by Sumit Ghosh

Python

# Python3 program to find all distinct characters
# in a String
  
MAX_CHAR = 256
  
# Function to prdistinct characters in
# given Str[]
def printDistinct(Str):
  
    n = len(Str)
  
    # count[x] is going to store count of
    # character 'x' in Str. If x is not present,
    # then it is going to store 0.
    count = [0 for i in range(MAX_CHAR)]
  
    # index[x] is going to store index of character
    # 'x' in Str. If x is not present or x is
    # more than once, then it is going to store a value
    # (for example, length of String) that cannot be
    # a valid index in Str[]
    index = [n for i in range(MAX_CHAR)]
  
  
    # Traverse the input String
    for i in range(n):
          
        # Find current character and increment its
        # count
        x = ord(Str[i])
        count[x] += 1
  
        # If this is first occurrence, then set value
        # in index as index of it.
        if (count[x] == 1 and x !=' '):
            index[x] = i
  
        # If character repeats, then remove it from
        # index[]
        if (count[x] == 2):
            index[x] = n
  
  
    # Since size of index is constant, below operations
    # take constant time.
    index=sorted(index)
  
    for i in range(MAX_CHAR):
        if index[i] == n:
            break
        print(Str[index[i]],end="")
  
# Driver code
  
Str = "GeeksforGeeks"
printDistinct(Str)
  
# This code is contributed by mohit kumar 29

C＃

// C# program to print distinct characters of 
// a string.
using System;
  
public class GFG {
      
    static int MAX_CHAR = 256;
      
    // Function to print distinct characters in
    // given string str[]
    static void printDistinct(string str)
    {
        int n = str.Length;
      
        // count[x] is going to store count of
        // character 'x' in str. If x is not 
        // present, then it is going to store 0.
        int []count = new int[MAX_CHAR];
      
        // index[x] is going to store index of 
        // character 'x' in str. If x is not 
        // present or x is more than once, then
        // it is going to store a value (for 
        // example, length of string) that 
        // cannot be a valid index in str[]
        int []index = new int[MAX_CHAR];
      
        // Initialize counts of all characters 
        // and indexes of distinct characters.
        for (int i = 0; i < MAX_CHAR; i++)
        {
            count[i] = 0;
              
            // A value more than any index
            // in str[]
            index[i] = n;
                          
        }
      
        // Traverse the input string
        for (int i = 0; i < n; i++)
        {
            // Find current character and 
            // increment its count
            char x = str[i];
            ++count[x];
      
            // If this is first occurrence, then
            // set value in index as index of it.
            if (count[x] == 1 && x !=' ')
                index[x] = i;
      
            // If character repeats, then remove
            // it from index[]
            if (count[x] == 2)
                index[x] = n;
        }
      
        // Since size of index is constant, below 
        // operations take constant time.
        Array.Sort(index);
          
        for (int i = 0; i < MAX_CHAR && 
                              index[i] != n; i++)
        Console.Write(str[index[i]]);
    }
      
    // Driver code
    public static void Main()
    {
        string str = "GeeksforGeeks";
        printDistinct(str);
    }
}
  
// This code is contributed by nitin mittal.

输出：

for