Daný vstupný reťazec a vzor skontrolujú, či znaky vo vstupnom reťazci majú rovnaké poradie, aké určujú znaky prítomné vo vzore. Predpokladajme, že vo vzore nebudú žiadne duplicitné znaky.
Ďalšie riešenie rovnakého problému je zverejnené tu .
Príklady:
Input: string = 'engineers rock' pattern = 'er'; Output: true All 'e' in the input string are before all 'r'. Input: string = 'engineers rock' pattern = 'egr'; Output: false There are two 'e' after 'g' in the input string. Input: string = 'engineers rock' pattern = 'gsr'; Output: false There are one 'r' before 's' in the input string.
Cieľom je zredukovať daný reťazec na daný vzor. Pre znaky uvedené vo vzore ponecháme v reťazci iba zodpovedajúce znaky. V novom reťazci vymažeme súvislé opakujúce sa znaky. Upravený reťazec by sa potom mal rovnať uvedenému vzoru. Nakoniec porovnáme upravený reťazec s daným vzorom a podľa toho vrátime hodnotu true alebo false.
Ilustrácia:
str = 'bfbaeadeacc' pat[] = 'bac' 1) Remove extra characters from str (characters that are not present in pat[] str = 'bbaaacc' [f e and d are removed] 3) Removed consecutive repeating occurrences of characters str = 'bac' 4) Since str is same as pat[] we return true
Nižšie je uvedená implementácia vyššie uvedených krokov.
// C++ code for the above approach #include #include using namespace std; bool followsPattern(string str string pattern) { // Insert all characters of pattern in a hash set unordered_set<char> patternSet; for (int i = 0; i < pattern.length(); i++) { patternSet.insert(pattern[i]); } // Build modified string (string with characters only from pattern are taken) string modifiedStr = str; for (int i = str.length() - 1; i >= 0; i--) { if (patternSet.find(str[i]) == patternSet.end()) { modifiedStr.erase(i 1); } } // Remove more than one consecutive occurrences of pattern characters from modified string for (int i = modifiedStr.length() - 1; i > 0; i--) { if (modifiedStr[i] == modifiedStr[i - 1]) { modifiedStr.erase(i 1); } } // After above modifications the length of modified string must be same as pattern length if (pattern.length() != modifiedStr.length()) { return false; } // And pattern characters must also be same as modified string characters for (int i = 0; i < pattern.length(); i++) { if (pattern[i] != modifiedStr[i]) { return false; } } return true; } int main() { string str = 'engineers rock'; string pattern = 'er'; cout << 'Expected: true Actual: ' << followsPattern(str pattern) << endl; str = 'engineers rock'; pattern = 'egr'; cout << 'Expected: false Actual: ' << followsPattern(str pattern) << endl; str = 'engineers rock'; pattern = 'gsr'; cout << 'Expected: false Actual: ' << followsPattern(str pattern) << endl; str = 'engineers rock'; pattern = 'eger'; cout << 'Expected: true Actual: ' << followsPattern(str pattern) << endl; return 0; } // This code is contributed by adityashatmfh
Java // Java program to check if characters of a string follow // pattern defined by given pattern. import java.util.*; public class OrderOfCharactersForPattern { public static boolean followsPattern(String str String pattern) { // Insert all characters of pattern in a hash set Set<Character> patternSet = neHashSet<>(); for (int i=0; i<pattern.length(); i++) patternSet.add(pattern.charAt(i)); // Build modified string (string with characters only from // pattern are taken) StringBuilder modifiedString = new StringBuilder(str); for (int i=str.length()-1; i>=0; i--) if (!patternSet.contains(modifiedString.charAt(i))) modifiedString.deleteCharAt(i); // Remove more than one consecutive occurrences of pattern // characters from modified string. for (int i=modifiedString.length()-1; i>0; i--) if (modifiedString.charAt(i) == modifiedString.charAt(i-1)) modifiedString.deleteCharAt(i); // After above modifications the length of modified string // must be same as pattern length if (pattern.length() != modifiedString.length()) return false; // And pattern characters must also be same as modified string // characters for (int i=0; i<pattern.length(); i++) if (pattern.charAt(i) != modifiedString.charAt(i)) return false; return true; } // Driver program int main() { String str = 'engineers rock'; String pattern = 'er'; System.out.println('Expected: true Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'egr'; System.out.println('Expected: false Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'gsr'; System.out.println('Expected: false Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'eger'; System.out.println('Expected: true Actual: ' + followsPattern(str pattern)); return 0; } }
Python3 # Python3 program to check if characters of # a string follow pattern defined by given pattern. def followsPattern(string pattern): # Insert all characters of pattern in a hash set patternSet = set() for i in range(len(pattern)): patternSet.add(pattern[i]) # Build modified string (string with characters # only from pattern are taken) modifiedString = string for i in range(len(string) - 1 -1 -1): if not modifiedString[i] in patternSet: modifiedString = modifiedString[:i] + modifiedString[i + 1:] # Remove more than one consecutive occurrences # of pattern characters from modified string. for i in range(len(modifiedString) - 1 0 -1): if modifiedString[i] == modifiedString[i - 1]: modifiedString = modifiedString[:i] + modifiedString[i + 1:] # After above modifications the length of # modified string must be same as pattern length if len(pattern) != len(modifiedString): return False # And pattern characters must also be same # as modified string characters for i in range(len(pattern)): if pattern[i] != modifiedString[i]: return False return True # Driver Code if __name__ == '__main__': string = 'engineers rock' pattern = 'er' print('Expected: true Actual:' followsPattern(string pattern)) string = 'engineers rock' pattern = 'egr' print('Expected: false Actual:' followsPattern(string pattern)) string = 'engineers rock' pattern = 'gsr' print('Expected: false Actual:' followsPattern(string pattern)) string = 'engineers rock' pattern = 'eger' print('Expected: true Actual:' followsPattern(string pattern)) # This code is contributed by # sanjeev2552
C# // C# program to check if characters of a string follow // pattern defined by given pattern. using System; using System.Collections.Generic; using System.Text; class GFG { public static bool followsPattern(String str String pattern) { // Insert all characters of pattern in a hash set HashSet<char> patternSet = new HashSet<char>(); for (int i = 0; i < pattern.Length; i++) patternSet.Add(pattern[i]); // Build modified string (string with characters // only from pattern are taken) StringBuilder modifiedString = new StringBuilder(str); for (int i = str.Length - 1; i >= 0; i--) if (!patternSet.Contains(modifiedString[i])) modifiedString.Remove(i 1); // Remove more than one consecutive occurrences of pattern // characters from modified string. for (int i = modifiedString.Length - 1; i > 0; i--) if (modifiedString[i] == modifiedString[i - 1]) modifiedString.Remove(i 1); // After above modifications the length of modified string // must be same as pattern length if (pattern.Length != modifiedString.Length) return false; // And pattern characters must also be same // as modified string characters for (int i = 0; i < pattern.Length; i++) if (pattern[i] != modifiedString[i]) return false; return true; } // Driver program public static void Main(String[] args) { String str = 'engineers rock'; String pattern = 'er'; Console.WriteLine('Expected: true Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'egr'; Console.WriteLine('Expected: false Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'gsr'; Console.WriteLine('Expected: false Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'eger'; Console.WriteLine('Expected: true Actual: ' + followsPattern(str pattern)); } } // This code is contributed by 29AjayKumar
JavaScript <script> // Javascript program to check if characters of a string follow // pattern defined by given pattern. function followsPattern(str pattern) { // Insert all characters of pattern in a hash set let patternSet = new Set(); for (let i=0; i<pattern.length; i++) patternSet.add(pattern[i]); // Build modified string (string with characters only from // pattern are taken) let modifiedString = (str).split(''); for (let i=str.length-1; i>=0; i--) if (!patternSet.has(modifiedString[i])) modifiedString.splice(i1); // Remove more than one consecutive occurrences of pattern // characters from modified string. for (let i=modifiedString.length-1; i>0; i--) if (modifiedString[i] == modifiedString[i-1]) modifiedString.splice(i1); // After above modifications the length of modified string // must be same as pattern length if (pattern.length != modifiedString.length) return false; // And pattern characters must also be same as modified string // characters for (let i=0; i<pattern.length; i++) if (pattern[i] != modifiedString[i]) return false; return true; } // Driver program let str = 'engineers rock'; let pattern = 'er'; document.write('Expected: true Actual: ' + followsPattern(str pattern)+'
'); str = 'engineers rock'; pattern = 'egr'; document.write('Expected: false Actual: ' + followsPattern(str pattern)+'
'); str = 'engineers rock'; pattern = 'gsr'; document.write('Expected: false Actual: ' + followsPattern(str pattern)+'
'); str = 'engineers rock'; pattern = 'eger'; document.write('Expected: true Actual: ' + followsPattern(str pattern)+'
'); // This code is contributed by rag2127 </script>
výstup:
Expected: true Actual: true Expected: false Actual: false Expected: false Actual: false Expected: true Actual: true
Časová zložitosť: Časová zložitosť vyššie uvedených implementácií je v skutočnosti O(mn + n^2), keďže na odstránenie znakov používame deleteCharAt(). Vyššie uvedené riešenie môžeme optimalizovať tak, aby fungovalo v lineárnom čase. Namiesto použitia deleteCharAr() môžeme vytvoriť prázdny reťazec a pridať doň iba požadované znaky.
StringBuilder sa používa na prácu so vstupným reťazcom. Je to preto, že StringBuilder je meniteľný, zatiaľ čo String je nemenný objekt. Vytvorenie nového reťazca zaberie O(n) priestor, takže priestor navyše je O(n).
Diskutovali sme o dvoch ďalších prístupoch k riešeniu tohto problému.
Skontrolujte, či reťazec nasleduje poradie znakov definovaných vzorom alebo nie | Set 1
Skontrolujte, či reťazec nasleduje poradie znakov definovaných vzorom alebo nie | Súprava 3