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Udskriv alle måder at bryde en streng i bracket -form

Givet en streng Find alle måder at bryde den givne streng i bracket -form. Vedlæg hver substring inden for en parentes.

partielle derivater i latex

Eksempler: 



Input : abc Output: (a)(b)(c) (a)(bc) (ab)(c) (abc) Input : abcd Output : (a)(b)(c)(d) (a)(b)(cd) (a)(bc)(d) (a)(bcd) (ab)(c)(d) (ab)(cd) (abc)(d) (abcd)

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Ideen er at bruge rekursion. Vi opretholder to parametre - indeks for den næste karakter, der skal behandles, og outputstrengen indtil videre. Vi starter fra indeks for næste karakter, der skal behandles tilsættende substring dannet af uforarbejdet streng til outputstrengen og recurse på den resterende streng, indtil vi behandler hele strengen. Vi bruger STD :: Substr til at danne outputstrengen. Substr (POS N) returnerer en substring af længde N, der starter ved position POS for den nuværende streng.

sql datatyper

Nedenfor diagram viser rekursionstræ til inputstreng 'ABC'. Hver knude på diagrammet viser behandlet streng (markeret med grøn) og uforarbejdet streng (markeret med rødt).



breakastring' title=

Nedenfor er implementeringen af ​​ovenstående idé

selen grundlæggende
C++
// C++ Program to find all combinations of Non- // overlapping substrings formed from given // string #include    using namespace std; // find all combinations of non-overlapping // substrings formed by input string str // index – index of the next character to // be processed // out - output string so far void findCombinations(string str int index string out) {  if (index == str.length())  cout << out << endl;  for (int i = index; i < str.length(); i++)  {  // append substring formed by str[index  // i] to output string  findCombinations(  str   i + 1  out + '(' + str.substr(index i + 1 - index)  + ')');  } } // Driver Code int main() {  // input string  string str = 'abcd';  findCombinations(str 0 '');  return 0; } 
Java
// Java program to find all combinations of Non- // overlapping substrings formed from given // string class GFG  {  // find all combinations of non-overlapping  // substrings formed by input string str  static void findCombinations(String str int index  String out)  {  if (index == str.length())  System.out.println(out);    for (int i = index; i < str.length(); i++)    // append substring formed by str[index  // i] to output string  findCombinations(str i + 1 out +  '(' + str.substring(index i+1) + ')' );  }    // Driver Code  public static void main (String[] args)   {  // input string  String str = 'abcd';  findCombinations(str 0 '');  } } // Contributed by Pramod Kumar 
Python3
# Python3 Program to find all combinations of Non- # overlapping substrings formed from given # string # find all combinations of non-overlapping # substrings formed by input string str # index – index of the next character to # be processed # out - output string so far def findCombinations(string index out): if index == len(string): print(out) for i in range(index len(string) 1): # append substring formed by str[index # i] to output string findCombinations(string i + 1 out + '(' + string[index:i + 1] + ')') # Driver Code if __name__ == '__main__': # input string string = 'abcd' findCombinations(string 0 '') # This code is contributed by # sanjeev2552 
C#
// C# program to find all combinations // of Non-overlapping substrings formed // from given string using System; class GFG {  // find all combinations of non-overlapping  // substrings formed by input string str  public static void  findCombinations(string str int index string @out)  {  if (index == str.Length) {  Console.WriteLine(@out);  }  for (int i = index; i < str.Length; i++) {  // append substring formed by  // str[index i] to output string  findCombinations(  str i + 1  @out + '('  + str.Substring(index (i + 1) - index)  + ')');  }  }  // Driver Code  public static void Main(string[] args)  {  // input string  string str = 'abcd';  findCombinations(str 0 '');  } } // This code is contributed by Shrikant13 
JavaScript
// Javascript program for the above approach // find all combinations of non-overlapping // substrings formed by input string str // index – index of the next character to // be processed // out - output string so far function findCombinations(string index out) {  if (index == string.length) {  console.log(out);  }  for (let i = index; i < string.length; i++) {  // append substring formed by str[index  // i] to output string  findCombinations(string i + 1 out + '(' + string.substring(index i + 1) + ')');  } } // Driver Code const string = 'abcd'; findCombinations(string 0 ''); // contributed by adityasharmadev01 

Produktion
(a)(b)(c)(d) (a)(b)(cd) (a)(bc)(d) (a)(bcd) (ab)(c)(d) (ab)(cd) (abc)(d) (abcd)

Tidskompleksitet: o (n2)
Hjælpeplads: o (n2)