查找字符串中最后一个不重复的字符

给定一个字符串str ，任务是找到其中的最后一个非重复字符。
例如，如果输入字符串是“GeeksForGeeks” ，那么输出应该是'r' ，如果输入字符串是“GeeksQuiz” ，那么输出应该是'z' 。如果没有非重复字符，则打印-1 。
例子：

Input: str = “GeeksForGeeks” 
Output: r 
‘r’ is the first character from the end which has frequency 1.
Input: str = “aabbcc” 
Output: -1 
All the characters of the given string have frequencies greater than 1. 
 

方法：创建一个频率数组，该数组将存储给定字符串中每个字符的频率。更新频率后，开始逐个字符遍历字符串，对于每个字符，如果当前字符的频率为 1，则这是最后字符。如果所有字符的频率都大于 1，则打印 -1。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Maximum distinct characters possible
const int MAX = 256;
 
// Function to return the last non-repeating character
static string lastNonRepeating(string str, int n)
{
 
    // To store the frequency of each of
    // the character of the given string
    int freq[MAX] = {0};
 
    // Update the frequencies
    for (int i = 0; i < n; i++)
        freq[str.at(i)]++;
     
 
    // Starting from the last character
    for (int i = n - 1; i >= 0; i--)
    {
 
        // Current character
        char ch = str.at(i);
 
        // If frequency of the current character is 1
        // then return the character
        if (freq[ch] == 1)
        {
            string res;
            res+=ch;
            return res;
        }
    }
 
    // All the characters of the
    // string are repeating
    return "-1";
}
 
// Driver code
int main()
{
    string str = "GeeksForGeeks";
    int n = str.size();
    cout<< lastNonRepeating(str, n);
    return 0;
}
 
// This code has been contributed by 29AjayKumar

Java

// Java implementation of the approach
public class GFG {
 
    // Maximum distinct characters possible
    static final int MAX = 256;
 
    // Function to return the last non-repeating character
    static String lastNonRepeating(String str, int n)
    {
 
        // To store the frequency of each of
        // the character of the given string
        int freq[] = new int[MAX];
 
        // Update the frequencies
        for (int i = 0; i < n; i++)
            freq[str.charAt(i)]++;
 
        // Starting from the last character
        for (int i = n - 1; i >= 0; i--) {
 
            // Current character
            char ch = str.charAt(i);
 
            // If frequency of the current character is 1
            // then return the character
            if (freq[ch] == 1)
                return ("" + ch);
        }
 
        // All the characters of the
        // string are repeating
        return "-1";
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "GeeksForGeeks";
        int n = str.length();
        System.out.println(lastNonRepeating(str, n));
    }
}

Python3

# Python3 implementation of the approach
 
# Maximum distinct characters possible
MAX = 256;
 
# Function to return the last non-repeating character
def lastNonRepeating(string, n) :
 
    # To store the frequency of each of
    # the character of the given string
    freq = [0]*MAX;
 
    # Update the frequencies
    for i in range(n) :
        freq[ord(string[i])] += 1;
 
    # Starting from the last character
    for i in range(n-1,-1,-1) :
 
        # Current character
        ch = string[i];
 
        # If frequency of the current character is 1
        # then return the character
        if (freq[ord(ch)] == 1) :
            return ("" + ch);
     
 
    # All the characters of the
    # string are repeating
    return "-1";
 
 
# Driver code
if __name__ == "__main__" :
     
    string = "GeeksForGeeks";
    n = len(string);
     
    print(lastNonRepeating(string, n));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
     
class GFG
{
 
    // Maximum distinct characters possible
    static readonly int MAX = 256;
 
    // Function to return the last non-repeating character
    static String lastNonRepeating(String str, int n)
    {
 
        // To store the frequency of each of
        // the character of the given string
        int []freq = new int[MAX];
 
        // Update the frequencies
        for (int i = 0; i < n; i++)
            freq[str[i]]++;
 
        // Starting from the last character
        for (int i = n - 1; i >= 0; i--)
        {
 
            // Current character
            char ch = str[i];
 
            // If frequency of the current character is 1
            // then return the character
            if (freq[ch] == 1)
                return ("" + ch);
        }
 
        // All the characters of the
        // string are repeating
        return "-1";
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String str = "GeeksForGeeks";
        int n = str.Length;
        Console.WriteLine(lastNonRepeating(str, n));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// JavaScript implementation of the approach
     
     // Maximum distinct characters possible
    let MAX = 256;
     
    // Function to return the last non-repeating character
    function lastNonRepeating(str,n)
    {
        // To store the frequency of each of
        // the character of the given string
        let freq = new Array(MAX);
          for(let i=0;i<MAX;i++)
        {
            freq[i]=0;
        }
        // Update the frequencies
        for (let i = 0; i < n; i++)
            freq[str[i].charCodeAt(0)]++;
   
        // Starting from the last character
        for (let i = n - 1; i >= 0; i--) {
   
            // Current character
            let ch = str[i];
   
            // If frequency of the current character is 1
            // then return the character
            if (freq[ch.charCodeAt(0)] == 1)
                return ("" + ch);
        }
   
        // All the characters of the
        // string are repeating
        return "-1";
    }
     
    // Driver code
    let str = "GeeksForGeeks";
    let n = str.length;
    document.write(lastNonRepeating(str, n));
 
 
// This code is contributed by rag2127
 

输出：

r