给定字符串的按字典顺序排列的最大子序列

给定一个包含小写字符的字符串str ，任务是找到按字典顺序排列的str的最大子序列。

例子：

Input: str = “abc”
Output: c
All possible sub-sequences are “a”, “ab”, “ac”, “b”, “bc” and “c”
and “c” is the largest among them (lexicographically)

Input: str = “geeksforgeeks”
Output: ss

方法：假设mx是字符串按字典顺序最大的字符。由于我们要按字典顺序排列最大的子序列，因此我们应包括所有出现的mx 。现在，在使用所有出现的字符之后，可以对剩余的字符串(即最后一次出现的mx之后的子字符串)重复相同的过程，依此类推，直到没有剩余的字符为止。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the lexicographically
// largest sub-sequence of s
string getSubSeq(string s, int n)
{
    string res = "";
    int cr = 0;
    while (cr < n) {
  
        // Get the max character from the string
        char mx = s[cr];
        for (int i = cr + 1; i < n; i++)
            mx = max(mx, s[i]);
        int lst = cr;
  
        // Use all the occurrences of the
        // current maximum character
        for (int i = cr; i < n; i++)
            if (s[i] == mx) {
                res += s[i];
                lst = i;
            }
  
        // Repeat the steps for the remaining string
        cr = lst + 1;
    }
    return res;
}
  
// Driver code
int main()
{
    string s = "geeksforgeeks";
    int n = s.length();
    cout << getSubSeq(s, n);
}

Java

// Java implementation of the approach
class GFG
{
  
    // Function to return the lexicographically
    // largest sub-sequence of s
    static String getSubSeq(String s, int n)
    {
        String res = "";
        int cr = 0;
        while (cr < n) 
        {
  
            // Get the max character from the String
            char mx = s.charAt(cr);
            for (int i = cr + 1; i < n; i++)
            {
                mx = (char) Math.max(mx, s.charAt(i));
            }
            int lst = cr;
  
            // Use all the occurrences of the
            // current maximum character
            for (int i = cr; i < n; i++) 
            {
                if (s.charAt(i) == mx) 
                {
                    res += s.charAt(i);
                    lst = i;
                }
            }
  
            // Repeat the steps for 
            // the remaining String
            cr = lst + 1;
        }
        return res;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        String s = "geeksforgeeks";
        int n = s.length();
        System.out.println(getSubSeq(s, n));
    }
}
  
// This code is contributed by Rajput-Ji

Python3

# Python 3 implementation of the approach
  
# Function to return the lexicographically
# largest sub-sequence of s
def getSubSeq(s, n):
    res = ""
    cr = 0
    while (cr < n):
          
        # Get the max character from 
        # the string
        mx = s[cr]
        for i in range(cr + 1, n):
            mx = max(mx, s[i])
        lst = cr
  
        # Use all the occurrences of the
        # current maximum character
        for i in range(cr,n):
            if (s[i] == mx):
                res += s[i]
                lst = i
  
        # Repeat the steps for the 
        # remaining string
        cr = lst + 1
      
    return res
  
# Driver code
if __name__ == '__main__':
    s = "geeksforgeeks"
    n = len(s)
    print(getSubSeq(s, n))
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach 
using System;
  
class GFG 
{ 
  
    // Function to return the lexicographically 
    // largest sub-sequence of s 
    static String getSubSeq(String s, int n) 
    { 
        String res = ""; 
        int cr = 0; 
        while (cr < n) 
        { 
  
            // Get the max character from 
            // the String 
            char mx = s[cr]; 
            for (int i = cr + 1; i < n; i++) 
            { 
                mx = (char) Math.Max(mx, s[i]); 
            } 
            int lst = cr; 
  
            // Use all the occurrences of the 
            // current maximum character 
            for (int i = cr; i < n; i++) 
            { 
                if (s[i] == mx) 
                { 
                    res += s[i]; 
                    lst = i; 
                } 
            } 
  
            // Repeat the steps for 
            // the remaining String 
            cr = lst + 1; 
        } 
        return res; 
    } 
  
    // Driver code 
    public static void Main(String[] args) 
    { 
        String s = "geeksforgeeks"; 
        int n = s.Length; 
        Console.WriteLine(getSubSeq(s, n)); 
    } 
} 
  
// This code is contributed by 29AjayKumar

输出：

ss

时间复杂度： O(N)，其中N是字符串的长度。