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Sorter en streng i den rækkefølge, der er defineret af en anden streng

Givet to strenge (af små bogstaver) et mønster og en streng. Opgaven er at sortere strenge efter den rækkefølge, der er defineret af mønsteret. Det kan antages, at mønsteret har alle tegn i strengen, og at alle tegn i mønsteret kun vises én gang.

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Eksempler:  

Input : pat = 'bca' str = 'abc' Output : str = 'bca' Input : pat = 'bxyzca' str = 'abcabcabc' Output : str = 'bbbcccaaa' Input : pat = 'wcyuogmlrdfphitxjakqvzbnes' str = 'jcdokai' Output : str = 'codijak'

Fremgangsmåde 1: Ideen er først at tælle forekomster af alle tegn i str og gemme disse tællinger i et tællearray. Gå derefter gennem mønsteret fra venstre mod højre og for hvert tegn pat[i] se hvor mange gange det optræder i tællearray og kopier dette tegn disse mange gange til str.
Nedenfor er implementeringen af ​​ovenstående idé. 



Implementering:

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> 

Produktion
bca

Tidskompleksitet: O(m+n) hvor m er længden af ​​mønsteret og n er længden af ​​str.
Hjælpeplads:  O(1)  

Fremgangsmåde 2: 

Vi kan sende en komparator til funktionen sort() og sortere strengen efter mønsteret.

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 

Produktion
codijak


Tidskompleksitet: O(m+nlogn ) hvor m er længden af ​​mønsteret og n er længden af ​​str.
 Hjælpeplads:  O(1)

Øvelse : I ovenstående løsning er det antaget, at mønsteret har alle tegn af str. Overvej en modificeret version, hvor mønsteret måske ikke har alle tegn, og opgaven er at sætte alle resterende tegn (i strengen, men ikke i mønsteret) til sidst. De resterende tegn skal placeres i alfabetisk sorteret rækkefølge. Tip: I den anden løkke, når vi øger indekset og sætter tegnet i str, kan vi også reducere antallet på det tidspunkt. Og til sidst krydser vi tællearrayet for at placere de resterende tegn i alfabetisk sorteret rækkefølge.

 

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