logo

Sorteer een tekenreeks volgens de volgorde die door een andere tekenreeks is gedefinieerd

Gegeven twee strings (van kleine letters), een patroon en een string. De taak is om strings te sorteren volgens de volgorde die door het patroon is gedefinieerd. Er mag worden aangenomen dat het patroon alle karakters van de string bevat en dat alle karakters in het patroon slechts één keer voorkomen.

Voorbeelden:  

een lijst herhalen in Java
Input : pat = 'bca' str = 'abc' Output : str = 'bca' Input : pat = 'bxyzca' str = 'abcabcabc' Output : str = 'bbbcccaaa' Input : pat = 'wcyuogmlrdfphitxjakqvzbnes' str = 'jcdokai' Output : str = 'codijak'

Benadering 1: Het idee is om eerst alle tekens in str te tellen en deze tellingen op te slaan in een telarray. Doorloop vervolgens het patroon van links naar rechts en kijk voor elk teken pat[i] hoe vaak het voorkomt in de count-array en kopieer dit teken zo vaak naar str.
Hieronder vindt u de implementatie van het bovenstaande idee. 



Uitvoering:

C++
// C++ program to sort a string according to the // order defined by a pattern string #include    using namespace std; const int MAX_CHAR = 26; // Sort str according to the order defined by pattern. void sortByPattern(string& str string pat) {  // Create a count array store count of characters in str.  int count[MAX_CHAR] = { 0 };  // Count number of occurrences of each character  // in str.  for (int i = 0; i < str.length(); i++)  count[str[i] - 'a']++;  // Traverse the pattern and print every characters  // same number of times as it appears in str. This  // loop takes O(m + n) time where m is length of  // pattern and n is length of str.  int index = 0;  for (int i = 0; i < pat.length(); i++)  for (int j = 0; j < count[pat[i] - 'a']; j++)  str[index++] = pat[i]; } // Driver code int main() {  string pat = 'bca';  string str = 'abc';  sortByPattern(str pat);  cout << str;  return 0; } 
Java
// Java program to sort a string according to the // order defined by a pattern string class GFG {  static int MAX_CHAR = 26;  // Sort str according to the order defined by pattern.  static void sortByPattern(char[] str char[] pat)  {  // Create a count array store  // count of characters in str.  int count[] = new int[MAX_CHAR];  // Count number of occurrences of  // each character in str.  for (int i = 0; i < str.length; i++) {  count[str[i] - 'a']++;  }  // Traverse the pattern and print every characters  // same number of times as it appears in str. This  // loop takes O(m + n) time where m is length of  // pattern and n is length of str.  int index = 0;  for (int i = 0; i < pat.length; i++) {  for (int j = 0; j < count[pat[i] - 'a']; j++) {  str[index++] = pat[i];  }  }  }  // Driver code  public static void main(String args[])  {  char[] pat = 'bca'.toCharArray();  char[] str = 'abc'.toCharArray();  sortByPattern(str pat);  System.out.println(String.valueOf(str));  } } // This code has been contributed by 29AjayKumar 
Python3
# Python3 program to sort a string according to  # the order defined by a pattern string MAX_CHAR = 26 # Sort str according to the order defined by pattern. def sortByPattern(str pat): global MAX_CHAR # Create a count array store count # of characters in str. count = [0] * MAX_CHAR # Count number of occurrences of  # each character in str. for i in range (0 len(str)): count[ord(str[i]) - 97] += 1 # Traverse the pattern and print every characters # same number of times as it appears in str. This # loop takes O(m + n) time where m is length of # pattern and n is length of str. index = 0; str = '' for i in range (0 len(pat)): j = 0 while(j < count[ord(pat[i]) - ord('a')]): str += pat[i] j = j + 1 index += 1 return str # Driver code pat = 'bca' str = 'abc' print(sortByPattern(str pat)) # This code is contributed by ihritik 
C#
// C# program to sort a string according to the // order defined by a pattern string using System; class GFG {  static int MAX_CHAR = 26;  // Sort str according to the order defined by pattern.  static void sortByPattern(char[] str char[] pat)  {  // Create a count array store  // count of characters in str.  int[] count = new int[MAX_CHAR];  // Count number of occurrences of  // each character in str.  for (int i = 0; i < str.Length; i++) {  count[str[i] - 'a']++;  }  // Traverse the pattern and print every characters  // same number of times as it appears in str. This  // loop takes O(m + n) time where m is length of  // pattern and n is length of str.  int index = 0;  for (int i = 0; i < pat.Length; i++) {  for (int j = 0; j < count[pat[i] - 'a']; j++) {  str[index++] = pat[i];  }  }  }  // Driver code  public static void Main(String[] args)  {  char[] pat = 'bca'.ToCharArray();  char[] str = 'abc'.ToCharArray();  sortByPattern(str pat);  Console.WriteLine(String.Join('' str));  } } /* This code contributed by PrinciRaj1992 */ 
JavaScript
<script> // Javascript program to sort a string according to the // order defined by a pattern string let MAX_CHAR = 26; // Sort str according to the order defined by pattern. function sortByPattern(strpat) {  // Create a count array stor  // count of characters in str.  let count = new Array(MAX_CHAR);  for(let i = 0; i < MAX_CHAR; i++)  {  count[i] = 0;  }    // Count number of occurrences of  // each character in str.  for (let i = 0; i < str.length; i++) {  count[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;  }    // Traverse the pattern and print every characters  // same number of times as it appears in str. This  // loop takes O(m + n) time where m is length of  // pattern and n is length of str.  let index = 0;  for (let i = 0; i < pat.length; i++) {  for (let j = 0; j < count[pat[i].charCodeAt(0) - 'a'.charCodeAt(0)]; j++) {  str[index++] = pat[i];  }  } } // Driver code let pat = 'bca'.split(''); let str = 'abc'.split(''); sortByPattern(str pat); document.write((str).join('')); // This code is contributed by rag2127 </script> 

Uitvoer
bca

Tijdcomplexiteit: O(m+n) waarbij m de lengte van het patroon is en n de lengte van str.
Hulpruimte: O(1)  

Benadering 2: 

We kunnen een comparator doorgeven aan de functie sort() en de string sorteren volgens het patroon.

C++
#include    using namespace std; // Declaring a vector globally that stores which character // is occurring first vector<int> position(26 -1); //Comparator function bool cmp(char& char1 char& char2) {  return position[char1 - 'a'] < position[char2 - 'a']; } int main() {  // Pattern  string pat = 'wcyuogmlrdfphitxjakqvzbnes';  for (int i = 0; i < pat.length(); i++) {  if (position[pat[i] - 'a'] == -1)  position[pat[i] - 'a'] = i;  }  // String to be sorted  string str = 'jcdokai';  // Passing a comparator to sort function  sort(str.begin() str.end() cmp);  cout << str; } 
Java
import java.util.*; class Main {  // Declaring a list globally that stores which character is occurring first  static List<Integer> position = new ArrayList<>(Collections.nCopies(26 -1));  // Comparator function  static int cmp(char char1 char char2) {  if (position.get(char1 - 'a') < position.get(char2 - 'a')) {  return -1;  } else if (position.get(char1 - 'a') > position.get(char2 - 'a')) {  return 1;  } else {  return 0;  }  }  public static void main(String[] args) {  // Pattern  String pat = 'wcyuogmlrdfphitxjakqvzbnes';  for (int i = 0; i < pat.length(); i++) {  if (position.get(pat.charAt(i) - 'a') == -1) {  position.set(pat.charAt(i) - 'a' i);  }  }  // String to be sorted  String str = 'jcdokai';  // Passing a comparator to the sorted function  char[] charArr = str.toCharArray();  Arrays.sort(charArr new Comparator<Character>() {  public int compare(Character c1 Character c2) {  return cmp(c1 c2);  }  });  String sortedStr = new String(charArr);  System.out.println(sortedStr);  } } 
Python3
from typing import List from functools import cmp_to_key # Declaring a list globally that stores which character is occurring first position: List[int] = [-1] * 26 # Comparator function def cmp(char1: str char2: str) -> int: if position[ord(char1) - ord('a')] < position[ord(char2) - ord('a')]: return -1 elif position[ord(char1) - ord('a')] > position[ord(char2) - ord('a')]: return 1 else: return 0 if __name__ == '__main__': # Pattern pat = 'wcyuogmlrdfphitxjakqvzbnes' for i in range(len(pat)): if position[ord(pat[i]) - ord('a')] == -1: position[ord(pat[i]) - ord('a')] = i # String to be sorted str = 'jcdokai' # Passing a comparator to the sorted function sorted_str = sorted(str key=cmp_to_key(cmp)) print(''.join(sorted_str)) # This code is contributed by adityashatmfh 
JavaScript
<script> // Declaring a vector globally that stores which character // is occurring first let position = new Array(26).fill(-1); //Comparator function function cmp(char1 char2) {  return position[char1.charCodeAt(0) - 'a'.charCodeAt(0)] - position[char2.charCodeAt(0) - 'a'.charCodeAt(0)]; } // driver code // Pattern let pat = 'wcyuogmlrdfphitxjakqvzbnes'; for (let i = 0; i <br pat.length; i++) {  if (position[pat.charCodeAt(i) - 'a'.charCodeAt(0)] == -1)  position[pat.charCodeAt(i) - 'a'.charCodeAt(0)] = i; } // String to be sorted let str = 'jcdokai'; // Passing a comparator to sort function str = str.split('').sort(cmp).join(''); document.write(str'
'
); // This code is contributed by Shinjan Patra </script>
C#
// C# program for the above approach using System; using System.Collections.Generic; using System.Linq; class GFG {  // Declaring a list globally that stores which character is occurring first  static List<int> position = Enumerable.Repeat(-1 26).ToList();  // Comparator function  static int Cmp(char char1 char char2) {  if (position[char1 - 'a'] < position[char2 - 'a']) {  return -1;  } else if (position[char1 - 'a'] > position[char2 - 'a']) {  return 1;  } else {  return 0;  }  }  public static void Main() {  // Pattern  string pat = 'wcyuogmlrdfphitxjakqvzbnes';  for (int i = 0; i < pat.Length; i++) {  if (position[pat[i] - 'a'] == -1) {  position[pat[i] - 'a'] = i;  }  }  // String to be sorted  string str = 'jcdokai';  // Passing a comparator to the sorted function  char[] charArr = str.ToCharArray();  Array.Sort(charArr new Comparison<char>(Cmp));  string sortedStr = new string(charArr);  Console.WriteLine(sortedStr);  } } // This code is contributed by sdeadityasharma 

Uitvoer
codijak


Tijdcomplexiteit: O(m+nlogn ) waarbij m de lengte van het patroon is en n de lengte van str.
 Hulpruimte: O(1)

sorteer arraylist in Java

Oefening : In de bovenstaande oplossing wordt aangenomen dat het patroon alle tekens van str. Overweeg een aangepaste versie waarbij het patroon mogelijk niet alle tekens bevat en het de taak is om alle resterende tekens (in de string maar niet in het patroon) aan het einde te plaatsen. De resterende tekens moeten in alfabetisch gesorteerde volgorde worden geplaatst. Tip: in de tweede lus kunnen we bij het verhogen van de index en het plaatsen van het teken in str het aantal op dat moment ook verlagen. En ten slotte doorkruisen we de telreeks om de resterende tekens in alfabetisch gesorteerde volgorde te plaatsen.

 

Quiz maken