Givet en inmatningssträng och ett mönster, kontrollera om tecken i inmatningssträngen följer samma ordning som bestäms av tecken som finns i mönstret. Anta att det inte finns några dubbletter av tecken i mönstret.
En annan lösning på samma problem publiceras här .
Exempel:
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.
Tanken här är att reducera den givna strängen till det givna mönstret. För tecken som anges i mönstret behåller vi endast motsvarande tecken i strängen. I den nya strängen tar vi bort kontinuerligt upprepade tecken. Den modifierade strängen ska då vara lika med det givna mönstret. Slutligen jämför vi modifierad sträng med det givna mönstret och returnerar true of false i enlighet därmed.
Illustration:
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
Nedan är implementeringen av ovanstående steg.
// C++ code for the above approach #include
// 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 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# 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
<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>
Produktion:
stänga av utvecklarläget
Expected: true Actual: true Expected: false Actual: false Expected: false Actual: false Expected: true Actual: true
Tidskomplexitet: Tidskomplexiteten för ovanstående implementeringar är faktiskt O(mn + n^2) eftersom vi använder deleteCharAt() för att ta bort tecken. Vi kan optimera ovanstående lösning för att fungera i linjär tid. Istället för att använda deleteCharAr() kan vi skapa en tom sträng och bara lägga till nödvändiga tecken till den.
StringBuilder används för att arbeta på inmatningssträng. Detta beror på att StringBuilder är föränderligt medan String är ett oföränderligt objekt. För att skapa en ny sträng tar det O(n) utrymme så extra utrymme är O(n).
Vi har diskuterat ytterligare två metoder för att lösa detta problem.
Kontrollera om strängen följer teckenordningen som definieras av ett mönster eller inte | Set 1
Kontrollera om strängen följer teckenordningen som definieras av ett mönster eller inte | Set 3