删除第一对相邻的相似字符，直到可能为止

给定字符串Str ，任务是删除相似字符的第一个相邻对，直到可以。

注意：删除相邻字符得到一个新的字符串，然后再从新字符串中删除重复相邻并不断重复这个过程，直到所有类似的相邻字符对将被删除。

例子：

Input: str = “keexxllx”
Output: kx
Step 0: Remove ee to get “kxxllx”
Step 1: Remove xx to get “kllx”
Step 2: Remove ll to get “kx”

Input: str = “abbaca”
Output: ca

方法：
在C++中使用string的back()和pop_back()方法STL解决上述问题。迭代字符串中的每个字符，如果相邻字符相同，则使用pop_back()函数删除相邻字符。最后，返回最后一个字符串。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Function to remove adjacent duplicates
string removeDuplicates(string S)
{
    string ans = "";
  
    // Iterate for every character 
    // in the string
    for (auto it : S) {
  
        // If ans string is empty or its last
        // character does not match with the 
        // current character then append this 
        // character to the string
        if (ans.empty() or ans.back() != it)
            ans.push_back(it);
  
        // Matches with the previous one
        else if (ans.back() == it)
            ans.pop_back();
    }
  
    // Return the answer
    return ans;
}
  
// Driver Code
int main()
{
    string str = "keexxllx";
    cout << removeDuplicates(str);
}

Java

// Java implementation of the above approach
class GFG 
{
  
    // Function to remove adjacent duplicates
    static String removeDuplicates(String S)
    {
        String ans = "";
  
        // Iterate for every character
        // in the string
        for (int i = 0; i < S.length(); i++) 
        {
  
            // If ans string is empty or its last
            // character does not match with the
            // current character then append this
            // character to the string
            if (ans.isEmpty() || 
                ans.charAt(ans.length() - 1) != S.charAt(i))
                ans += S.charAt(i);
  
            // Matches with the previous one
            else if (ans.charAt(ans.length() - 1) == S.charAt(i))
                ans = ans.substring(0, ans.length() - 1);
        }
  
        // Return the answer
        return ans;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        String str = "keexxllx";
        System.out.println(removeDuplicates(str));
    }
}
  
// This code is contributed by
// sanjeev2552

Python3

# Python3 implementation of the above approach 
  
# Function to remove adjacent duplicates 
def removeDuplicates(S) : 
  
    ans = ""; 
  
    # Iterate for every character 
    # in the string 
    for it in S :
  
        # If ans string is empty or its last 
        # character does not match with the 
        # current character then append this 
        # character to the string 
        if (ans == "" or ans[-1] != it) :
            ans += it ; 
  
        # Matches with the previous one 
        elif (ans[-1] == it) :
            ans = ans[:-1]; 
  
    # Return the answer 
    return ans; 
  
  
# Driver Code 
if __name__ == "__main__" :
  
    string = "keexxllx"; 
    print(removeDuplicates(string)); 
      
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
      
class GFG 
{
  
    // Function to remove adjacent duplicates
    static String removeDuplicates(String S)
    {
        String ans = "";
  
        // Iterate for every character
        // in the string
        for (int i = 0; i < S.Length; i++) 
        {
  
            // If ans string is empty or its last
            // character does not match with the
            // current character then append this
            // character to the string
            if (ans == "" || 
                ans[ans.Length - 1] != S[i])
                ans += S[i];
  
            // Matches with the previous one
            else if (ans[ans.Length - 1] == S[i])
                ans = ans.Substring(0, ans.Length - 1);
        }
  
        // Return the answer
        return ans;
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        String str = "keexxllx";
        Console.WriteLine(removeDuplicates(str));
    }
}
  
// This code is contributed by Rajput-Ji

输出：

kx

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