有效地查找字符串的第一个重复字符，而无需在一次遍历中使用任何其他数据结构

实现节省空间的算法，以检查字符串的第一个重复字符，而无需在一次遍历中使用任何其他数据结构。不允许使用其他数据结构，例如count数组，hash等。

例子 ：

Input :  abcfdeacf
Output : char = a, index= 6

想法是使用整数变量，并使用其二进制表示形式的位来存储是否存在字符。通常，一个整数至少有32位，我们只需要存储26个字符的存在/不存在。

C++

// Efficiently check First repeated character
// in C++ program 
#include
using namespace std;
  
// Returns -1 if all characters of str are
// unique.
// Assumptions : (1) str contains only characters
//                 from 'a' to 'z'
//             (2) integers are stored using 32
//                 bits
int FirstRepeated(string str)
{
    // An integer to store presence/absence
    // of 26 characters using its 32 bits.
    int checker = 0;
  
    for (int i = 0; i < str.length(); ++i)
    {
        int val = (str[i]-'a');
  
        // If bit corresponding to current
        // character is already set
        if ((checker & (1 << val)) > 0)
            return i;
  
        // set bit in checker
        checker |= (1 << val);
    }
  
    return -1;
}
  
// Driver code
int main()
{
    string s = "abcfdeacf";
    int i=FirstRepeated(s);
    if (i!=-1)
        cout <<"Char = "<< s[i] << "   and Index = "<

---

Java

// Efficiently check First repeated character
// in Java program 
public class First_Repeated_char {
  
    static int FirstRepeated(String str)
    {
        // An integer to store presence/absence
        // of 26 characters using its 32 bits.
        int checker = 0;
       
        for (int i = 0; i < str.length(); ++i)
        {
            int val = (str.charAt(i)-'a');
       
            // If bit corresponding to current
            // character is already set
            if ((checker & (1 << val)) > 0)
                return i;
       
            // set bit in checker
            checker |= (1 << val);
        }
       
        return -1;
    }
       
    // Driver code
    public static void main(String args[])
    {
        String s = "abcfdeacf";
        int i=FirstRepeated(s);
        if (i!=-1)
           System.out.println("Char = "+ s.charAt(i) + "   and Index = "+i);
        else
            System.out.println( "No repeated Char");
    }
}
// This code is contributed by Sumit Ghosh

Python

# Efficiently check First repeated character
# in Python 
  
# Returns -1 if all characters of str are
# unique.
# Assumptions : (1) str contains only characters
#                 from 'a' to 'z'
##             (2) integers are stored using 32
##                 bits
def FirstRepeated(string):
      
    # An integer to store presence/absence
    # of 26 characters using its 32 bits.
    checker = 0
   
    pos = 0
    for i in string:
        val = ord(i) - ord('a');
   
        # If bit corresponding to current
        # character is already set
        if ((checker & (1 << val)) > 0):
            return pos
   
        # set bit in checker
        checker |= (1 << val)
        pos += 1
   
    return -1
   
# Driver code
string = "abcfdeacf"
i = FirstRepeated(string)
if i != -1:
    print "Char = ", string[i], " and Index = ", i;
else:
    print "No repeated Char"
  
# This code is contributed by Sachin Bisht

C#

// C# program to Efficiently 
// check First repeated character
using System;
  
public class First_Repeated_char {
  
    static int FirstRepeated(string str)
    {
        // An integer to store presence/absence
        // of 26 characters using its 32 bits.
        int checker = 0;
      
        for (int i = 0; i < str.Length; ++i)
        {
            int val = (str[i]-'a');
      
            // If bit corresponding to current
            // character is already set
            if ((checker & (1 << val)) > 0)
                return i;
      
            // set bit in checker
            checker |= (1 << val);
        }
      
        return -1;
    }
      
    // Driver code
    public static void Main()
    {
        string s = "abcfdeacf";
        int i=FirstRepeated(s);
        if (i!=-1)
           Console.WriteLine("Char = " + s[i] +
                          " and Index = " + i);
        else
            Console.WriteLine( "No repeated Char");
    }
}
  
// This code is contributed by vt_m.

PHP

 0)
            return $i;
  
        // set bit in checker
        $checker |= (1 << $val);
    }
  
    return -1;
}
  
// Driver code
$s = "abcfdeacf";
$i=FirstRepeated($s);
if ($i!=-1)
    echo "Char = " . $s[$i] .
         " and Index = " . $i;
else
    echo "No repeated Char";
  
// This code is contributed by ita_c
?>

输出：

Char = a   and Index = 6

时间复杂度： O(n)
辅助空间： O(1)