打印输入字符串中的所有重复项

编写一个高效的程序来打印输入字符串中的所有重复项及其计数

方法一：使用散列
算法：让输入字符串为“geeksforgeeks”
1：从输入字符串构造字符计数数组。
计数['e'] = 4
计数['g'] = 2
计数['k'] = 2
……
2：打印构造数组中值大于 1 的所有索引。
解决方案

C++14

// C++ program to count all duplicates
// from string using hashing
#include 
using namespace std;
# define NO_OF_CHARS 256
 
class gfg
{
    public :
     
    /* Fills count array with
    frequency of characters */
    void fillCharCounts(char *str, int *count)
    {
        int i;
        for (i = 0; *(str + i); i++)
        count[*(str + i)]++;
    }
 
    /* Print duplicates present
    in the passed string */
    void printDups(char *str)
    {
         
        // Create an array of size 256 and fill
        // count of every character in it
        int *count = (int *)calloc(NO_OF_CHARS,
                                      sizeof(int));
        fillCharCounts(str, count);
 
        // Print characters having count more than 0
        int i;
        for (i = 0; i < NO_OF_CHARS; i++)
        if(count[i] > 1)
            printf("%c, count = %d \n", i, count[i]);
 
        free(count);
    }
};
 
/* Driver code*/
int main()
{
    gfg g ;
    char str[] = "test string";
    g.printDups(str);
    //getchar();
    return 0;
}
 
// This code is contributed by SoM15242

C

// C program to count all duplicates
// from string using hashing
# include 
# include 
# define NO_OF_CHARS 256
 
/* Fills count array with
   frequency of characters */
void fillCharCounts(char *str, int *count)
{
   int i;
   for (i = 0; *(str+i);  i++)
      count[*(str+i)]++;
}
 
/* Print duplicates present
   in the passed string */
void printDups(char *str)
{
   
  // Create an array of size 256 and
  // fill count of every character in it
  int *count = (int *)calloc(NO_OF_CHARS,
                             sizeof(int));
  fillCharCounts(str, count);
 
  // Print characters having count more than 0
  int i;
  for (i = 0; i < NO_OF_CHARS; i++)
    if(count[i] > 1)
        printf("%c,  count = %d \n", i,  count[i]);
 
  free(count);
}
 
/* Driver program to test to print printDups*/
int main()
{
    char str[] = "test string";
    printDups(str);
    getchar();
    return 0;
}

Java

// Java program to count all
// duplicates from string using
// hashing
 
public class GFG {
    static final int NO_OF_CHARS = 256;
 
    /* Fills count array with
       frequency of characters */
    static void fillCharCounts(String str,
                                   int[] count)
    {
        for (int i = 0; i < str.length(); i++)
            count[str.charAt(i)]++;
    }
 
    /* Print duplicates present
      in the passed string */
    static void printDups(String str)
    {
        // Create an array of size
        // 256 and fill count of
        // every character in it
        int count[] = new int[NO_OF_CHARS];
        fillCharCounts(str, count);
 
        for (int i = 0; i < NO_OF_CHARS; i++)
            if (count[i] > 1)
                System.out.println((char)(i) +
                          ", count = " + count[i]);
    }
 
    // Driver Method
    public static void main(String[] args)
    {
        String str = "test string";
        printDups(str);
    }
}

Python

# Python program to count all
# duplicates from string using hashing
NO_OF_CHARS = 256
 
# Fills count array with
# frequency of characters
def fillCharCounts(string, count):
    for i in string:
        count[ord(i)] += 1
    return count
 
# Print duplicates present
# in the passed string
def printDups(string):
     
    # Create an array of size 256
    # and fill count of every character in it
    count = [0] * NO_OF_CHARS
    count = fillCharCounts(string,count)
 
    # Utility Variable
    k = 0
 
    # Print characters having
    # count more than 0
    for i in count:
        if int(i) > 1:
            print chr(k) + ", count = " + str(i)
        k += 1
 
# Driver program to test the above function
string = "test string"
print printDups(string)
 
# This code is contributed by Bhavya Jain

C#

// C# program to count all duplicates
// from string using hashing
using System;
 
class GFG
{
     
    static int NO_OF_CHARS = 256;
     
    /* Fills count array with
       frequency of characters */
    static void fillCharCounts(String str,
                                 int[] count)
    {
        for (int i = 0; i < str.Length; i++)
            count[str[i]]++;
    }
     
    /* Print duplicates present in
    the passed string */
    static void printDups(String str)
    {
         
        // Create an array of size 256 and
        // fill count of every character in it
        int []count = new int[NO_OF_CHARS];
        fillCharCounts(str, count);
     
        for (int i = 0; i < NO_OF_CHARS; i++)
            if(count[i] > 1)
                Console.WriteLine((char)i + ", " +
                              "count = " + count[i]);
    }
     
    // Driver Method
    public static void Main()
    {
        String str = "test string";
        printDups(str);
    }
}
 
// This code is contributed by Sam007

PHP

 1)
            echo chr($i) . ", " ."count = " .
                         ($count[$i]) . "\n";
}
 
// Print duplicates present
// in the passed string
function printDups($str)
{
    // Create an array of size
    // 256 and fill count of
    // every character in it
    $count = array();
    for ($i = 0; $i < 256; $i++)
    $count[$i] = 0;
    fillCharCounts($str, $count);
 
     
}
 
// Driver Code
$str = "test string";
printDups($str);
     
// This code is contributed by Sam007
?>

Javascript

    // Javascript program to count all duplicates
    // from string using hashing
     
    let NO_OF_CHARS = 256;
      
    /* Print duplicates present in
    the passed string */
    function printDups(str)
    {
          
        // Create an array of size 256 and
        // fill count of every character in it
        let count = new Array(NO_OF_CHARS);
        count.fill(0);
          
        for (let i = 0; i < str.length; i++)
            count[str[i].charCodeAt()]++;
             
        for (let i = 0; i < NO_OF_CHARS; i++)
        {
            if(count[i] > 1)
            {
                document.write(String.fromCharCode(i) + ", " +
                "count = " + count[i] + "</br>");
            }
        }    
    }
     
    let str = "test string";
      printDups(str);
         

输出

s, count = 2 
t, count = 3 

时间复杂度： O(n)，其中 n = 传递的字符串长度

空间复杂度：O(NO_OF_CHARS)

注意：散列涉及每次使用固定大小的数组，无论字符串是什么。

例如，str = “aaaaaaaaaa”。

str 使用大小为 256 的数组，总大小 (256) 中只有 1 个块将用于存储 str 中 'a' 出现的次数（即 count['a'] = 10）。

剩余 256 – 1 = 255 个块未使用。

因此，对于这种情况，空间复杂度可能很高。因此，为了避免任何差异并提高空间复杂性，地图通常比长数组更受欢迎。

方法 2：使用地图

方法：该方法与方法 1中讨论的方法相同，但使用映射来存储计数。

解决方案：

C++

// C++ program to count all duplicates
// from string using maps
#include 
using namespace std;
void printDups(string str)
{
    map count;
    for (int i = 0; i < str.length(); i++) {
        count[str[i]]++;
    }
 
    for (auto it : count) {
        if (it.second > 1)
            cout << it.first << ", count = " << it.second
                 << "\n";
    }
}
/* Driver code*/
int main()
{
    string str = "test string";
    printDups(str);
    return 0;
}
// This code is contributed by yashbeersingh42

Java

// Java program to count all duplicates
// from string using maps
import java.util.*;
 
class GFG {
 
    static void printDups(String str)
    {
        HashMap count = new HashMap<>();
        /*Store duplicates present
        in the passed string */
        for (int i = 0; i < str.length(); i++) {
            if (!count.containsKey(str.charAt(i)))
                count.put(str.charAt(i), 1);
            else
                count.put(str.charAt(i),
                          count.get(str.charAt(i)) + 1);
        }
 
        /*Print duplicates in sorted order*/
        for (Map.Entry mapElement : count.entrySet()) {
            char key = (char)mapElement.getKey();
            int value = ((int)mapElement.getValue());
 
            if (value > 1)
                System.out.println(key
                                   + ", count = " + value);
        }
    }
    // Driver code
    public static void main(String[] args)
    {
        String str = "test string";
        printDups(str);
    }
}
// This code is contributed by yashbeersingh42

Python3

# Python 3 program to count all duplicates
# from string using maps
from collections import defaultdict
 
def printDups(st):
 
    count = defaultdict(int)
    for i in range(len(st)):
        count[st[i]] += 1
 
    for it in count:
        if (count[it] > 1):
            print(it, ", count = ", count[it])
 
# Driver code
if __name__ == "__main__":
 
    st = "test string"
    printDups(st)
 
    # This code is contributed by ukasp.

C＃

// C# program to count all duplicates
// from string using maps
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG{
 
static void printDups(String str)
{
    Dictionary count = new Dictionary();
 
    for(int i = 0; i < str.Length; i++)
    {
        if (count.ContainsKey(str[i]))
            count[str[i]]++;
        else
            count[str[i]] = 1;
    }
 
    foreach(var it in count.OrderBy(key => key.Value))
    {
        if (it.Value > 1)
            Console.WriteLine(it.Key + ", count = " +
                              it.Value);
    }
}
 
// Driver code
static public void Main()
{
    String str = "test string";
    printDups(str);
}
}
 
// This code is contributed by shubhamsingh10

Javascript

//Javascript Implementation
//  to count all duplicates
// from string using maps
 
function printDups(str)
{
    var count = {};
    for (var i = 0; i < str.length; i++) {
        count[str[i]] = 0;
    }
     
    for (var i = 0; i < str.length; i++) {
        count[str[i]]++;
    }
 
    for (var it in count) {
        if (count[it] > 1)
            document.write(it + ", count = " + count[it] + "<br>");
    }
}
/* Driver code*/
var str = "test string";
printDups(str);
 
// This code is contributed by shubhamsingh10

输出

s, count = 2
t, count = 3

时间复杂度：O(N log N)，其中 N = 传递的字符串长度，通常需要 logN 时间将元素插入映射中。

空间复杂度：O(K)，其中 K = 地图的大小 ( 0<=K<=input_string_length )。

C++

// C++ program to count all duplicates
// from string using maps
#include 
using namespace std;
void printDups(string str)
{
    unordered_map count;
    for (int i = 0; i < str.length(); i++) {
        count[str[i]]++;  //increase the count of characters by 1
    }
  
    for (auto it : count) {   //iterating through the unordered map
        if (it.second > 1)   //if the count of characters is greater then 1 then duplicate found
            cout << it.first << ", count = " << it.second
                 << "\n";
    }
}
/* Driver code*/
int main()
{
    string str = "test string";
    printDups(str);
    return 0;
}

输出

s, count = 2
t, count = 3

Time Complexity:

O(N), where N = length of the string passed and it takes O(1) time to insert and access any element in an unordered map

Space Complexity:

 O(K), where K = size of the map (0<=K<=input_string_length).