Stapnummers
Probeer het eens op GfG Practice 
#practiceLinkDiv {weergave: geen! belangrijk; }
Uitvoer
#practiceLinkDiv {weergave: geen! belangrijk; } Gegeven twee gehele getallen 'n' en 'm', zoek dan alle stapgetallen binnen bereik [n m]. Er wordt een nummer gebeld stapnummer als alle aangrenzende cijfers een absoluut verschil van 1 hebben. 321 is een stapnummer, terwijl 421 dat niet is.
Voorbeelden:
Input : n = 0 m = 21 Output : 0 1 2 3 4 5 6 7 8 9 10 12 21 Input : n = 10 m = 15 Output : 10 12Recommended Practice Getallen met één absoluut verschil Probeer het!
Methode 1: Brute force-aanpak
Bij deze methode wordt een brute force-aanpak gebruikt om alle gehele getallen van n tot m te doorlopen en te controleren of het een stapsgewijs getal is.
// A C++ program to find all the Stepping Number in [n m] #include using namespace std ; // This function checks if an integer n is a Stepping Number bool isStepNum ( int n ) { // Initialize prevDigit with -1 int prevDigit = -1 ; // Iterate through all digits of n and compare difference // between value of previous and current digits while ( n ) { // Get Current digit int curDigit = n % 10 ; // Single digit is consider as a // Stepping Number if ( prevDigit == -1 ) prevDigit = curDigit ; else { // Check if absolute difference between // prev digit and current digit is 1 if ( abs ( prevDigit - curDigit ) != 1 ) return false ; } prevDigit = curDigit ; n /= 10 ; } return true ; } // A brute force approach based function to find all // stepping numbers. void displaySteppingNumbers ( int n int m ) { // Iterate through all the numbers from [NM] // and check if it’s a stepping number. for ( int i = n ; i <= m ; i ++ ) if ( isStepNum ( i )) cout < < i < < ' ' ; } // Driver program to test above function int main () { int n = 0 m = 21 ; // Display Stepping Numbers in // the range [n m] displaySteppingNumbers ( n m ); return 0 ; }
Java // A Java program to find all the Stepping Number in [n m] class Main { // This Method checks if an integer n // is a Stepping Number public static boolean isStepNum ( int n ) { // Initialize prevDigit with -1 int prevDigit = - 1 ; // Iterate through all digits of n and compare // difference between value of previous and // current digits while ( n > 0 ) { // Get Current digit int curDigit = n % 10 ; // Single digit is consider as a // Stepping Number if ( prevDigit != - 1 ) { // Check if absolute difference between // prev digit and current digit is 1 if ( Math . abs ( curDigit - prevDigit ) != 1 ) return false ; } n /= 10 ; prevDigit = curDigit ; } return true ; } // A brute force approach based function to find all // stepping numbers. public static void displaySteppingNumbers ( int n int m ) { // Iterate through all the numbers from [NM] // and check if it is a stepping number. for ( int i = n ; i <= m ; i ++ ) if ( isStepNum ( i )) System . out . print ( i + ' ' ); } // Driver code public static void main ( String args [] ) { int n = 0 m = 21 ; // Display Stepping Numbers in the range [nm] displaySteppingNumbers ( n m ); } }
Python3 # A Python3 program to find all the Stepping Number in [n m] # This function checks if an integer n is a Stepping Number def isStepNum ( n ): # Initialize prevDigit with -1 prevDigit = - 1 # Iterate through all digits of n and compare difference # between value of previous and current digits while ( n ): # Get Current digit curDigit = n % 10 # Single digit is consider as a # Stepping Number if ( prevDigit == - 1 ): prevDigit = curDigit else : # Check if absolute difference between # prev digit and current digit is 1 if ( abs ( prevDigit - curDigit ) != 1 ): return False prevDigit = curDigit n //= 10 return True # A brute force approach based function to find all # stepping numbers. def displaySteppingNumbers ( n m ): # Iterate through all the numbers from [NM] # and check if it’s a stepping number. for i in range ( n m + 1 ): if ( isStepNum ( i )): print ( i end = ' ' ) # Driver code if __name__ == '__main__' : n m = 0 21 # Display Stepping Numbers in # the range [n m] displaySteppingNumbers ( n m ) # This code is contributed by mohit kumar 29
C# // A C# program to find all // the Stepping Number in [n m] using System ; class GFG { // This Method checks if an // integer n is a Stepping Number public static bool isStepNum ( int n ) { // Initialize prevDigit with -1 int prevDigit = - 1 ; // Iterate through all digits // of n and compare difference // between value of previous // and current digits while ( n > 0 ) { // Get Current digit int curDigit = n % 10 ; // Single digit is considered // as a Stepping Number if ( prevDigit != - 1 ) { // Check if absolute difference // between prev digit and current // digit is 1 if ( Math . Abs ( curDigit - prevDigit ) != 1 ) return false ; } n /= 10 ; prevDigit = curDigit ; } return true ; } // A brute force approach based // function to find all stepping numbers. public static void displaySteppingNumbers ( int n int m ) { // Iterate through all the numbers // from [NM] and check if it is // a stepping number. for ( int i = n ; i <= m ; i ++ ) if ( isStepNum ( i )) Console . Write ( i + ' ' ); } // Driver code public static void Main () { int n = 0 m = 21 ; // Display Stepping Numbers // in the range [nm] displaySteppingNumbers ( n m ); } } // This code is contributed by nitin mittal.
JavaScript < script > // A Javascript program to find all the Stepping Number in [n m] // This function checks if an integer n is a Stepping Number function isStepNum ( n ) { // Initialize prevDigit with -1 let prevDigit = - 1 ; // Iterate through all digits of n and compare difference // between value of previous and current digits while ( n > 0 ) { // Get Current digit let curDigit = n % 10 ; // Single digit is consider as a // Stepping Number if ( prevDigit == - 1 ) prevDigit = curDigit ; else { // Check if absolute difference between // prev digit and current digit is 1 if ( Math . abs ( prevDigit - curDigit ) != 1 ) return false ; } prevDigit = curDigit ; n = parseInt ( n / 10 10 ); } return true ; } // A brute force approach based function to find all // stepping numbers. function displaySteppingNumbers ( n m ) { // Iterate through all the numbers from [NM] // and check if it’s a stepping number. for ( let i = n ; i <= m ; i ++ ) if ( isStepNum ( i )) document . write ( i + ' ' ); } let n = 0 m = 21 ; // Display Stepping Numbers in // the range [n m] displaySteppingNumbers ( n m ); // This code is contributed by mukesh07. < /script>
Uitvoer
0 1 2 3 4 5 6 7 8 9 10 12 21
Methode 2: BFS/DFS gebruiken
Het idee is om gebruik te maken van een Breedte eerste zoekopdracht / Diepte eerste zoekopdracht doortocht.
Hoe de grafiek opbouwen?
Elk knooppunt in de grafiek vertegenwoordigt een stapsgewijs getal; er zal een gerichte rand zijn van een knooppunt U naar V als V kan worden getransformeerd vanuit U. (U en V zijn stapnummers) Een stapnummer V kan op de volgende manier worden getransformeerd vanuit U.
laatstecijfer verwijst naar het laatste cijfer van U (d.w.z. U % 10)
Een aangrenzend nummer V kan zijn:
- U*10 + laatstecijfer + 1 (Buurman A)
- U*10 + laatstecijfer – 1 (Buurman B)
Door bovenstaande bewerkingen toe te passen wordt een nieuw cijfer toegevoegd aan U. Het is ofwel laatstecijfer-1 of laatstecijfer+1, zodat het nieuwe getal V gevormd uit U ook een stapnummer is.
Daarom zal elk knooppunt maximaal 2 aangrenzende knooppunten hebben.
Randgevallen: Wanneer het laatste cijfer van U is of 9
- Elk getal met één cijfer wordt beschouwd als een stapsgewijs getal, dus bfs-traversal voor elk cijfer geeft alle stapnummers vanaf dat cijfer.
- Voer een bfs/dfs-traversal uit voor alle getallen uit [09].
Wat zal het bron-/startknooppunt zijn?
Opmerking: Voor knooppunt 0 is het niet nodig om buren te verkennen tijdens de BFS-traversal, aangezien dit zal leiden tot 01 012 010 en deze zullen worden gedekt door de BFS-traversal vanaf knooppunt 1.
Voorbeeld om alle stapnummers van 0 tot 21 te vinden
-> 0 is a stepping Number and it is in the range so display it. -> 1 is a Stepping Number find neighbors of 1 i.e. 10 and 12 and push them into the queue How to get 10 and 12? Here U is 1 and last Digit is also 1 V = 10 + 0 = 10 ( Adding lastDigit - 1 ) V = 10 + 2 = 12 ( Adding lastDigit + 1 ) Then do the same for 10 and 12 this will result into 101 123 121 but these Numbers are out of range. Now any number transformed from 10 and 12 will result into a number greater than 21 so no need to explore their neighbors. -> 2 is a Stepping Number find neighbors of 2 i.e. 21 23. -> 23 is out of range so it is not considered as a Stepping Number (Or a neighbor of 2) The other stepping numbers will be 3 4 5 6 7 8 9.
Op BFS gebaseerde oplossing:
C++ // A C++ program to find all the Stepping Number from N=n // to m using BFS Approach #include using namespace std ; // Prints all stepping numbers reachable from num // and in range [n m] void bfs ( int n int m int num ) { // Queue will contain all the stepping Numbers queue < int > q ; q . push ( num ); while ( ! q . empty ()) { // Get the front element and pop from the queue int stepNum = q . front (); q . pop (); // If the Stepping Number is in the range // [n m] then display if ( stepNum <= m && stepNum >= n ) cout < < stepNum < < ' ' ; // If Stepping Number is 0 or greater than m // no need to explore the neighbors if ( num == 0 || stepNum > m ) continue ; // Get the last digit of the currently visited // Stepping Number int lastDigit = stepNum % 10 ; // There can be 2 cases either digit to be // appended is lastDigit + 1 or lastDigit - 1 int stepNumA = stepNum * 10 + ( lastDigit - 1 ); int stepNumB = stepNum * 10 + ( lastDigit + 1 ); // If lastDigit is 0 then only possible digit // after 0 can be 1 for a Stepping Number if ( lastDigit == 0 ) q . push ( stepNumB ); //If lastDigit is 9 then only possible //digit after 9 can be 8 for a Stepping //Number else if ( lastDigit == 9 ) q . push ( stepNumA ); else { q . push ( stepNumA ); q . push ( stepNumB ); } } } // Prints all stepping numbers in range [n m] // using BFS. void displaySteppingNumbers ( int n int m ) { // For every single digit Number 'i' // find all the Stepping Numbers // starting with i for ( int i = 0 ; i <= 9 ; i ++ ) bfs ( n m i ); } //Driver program to test above function int main () { int n = 0 m = 21 ; // Display Stepping Numbers in the // range [nm] displaySteppingNumbers ( n m ); return 0 ; }
Java // A Java program to find all the Stepping Number in // range [n m] import java.util.* ; class Main { // Prints all stepping numbers reachable from num // and in range [n m] public static void bfs ( int n int m int num ) { // Queue will contain all the stepping Numbers Queue < Integer > q = new LinkedList < Integer > (); q . add ( num ); while ( ! q . isEmpty ()) { // Get the front element and pop from // the queue int stepNum = q . poll (); // If the Stepping Number is in // the range [nm] then display if ( stepNum <= m && stepNum >= n ) { System . out . print ( stepNum + ' ' ); } // If Stepping Number is 0 or greater // then m no need to explore the neighbors if ( stepNum == 0 || stepNum > m ) continue ; // Get the last digit of the currently // visited Stepping Number int lastDigit = stepNum % 10 ; // There can be 2 cases either digit // to be appended is lastDigit + 1 or // lastDigit - 1 int stepNumA = stepNum * 10 + ( lastDigit - 1 ); int stepNumB = stepNum * 10 + ( lastDigit + 1 ); // If lastDigit is 0 then only possible // digit after 0 can be 1 for a Stepping // Number if ( lastDigit == 0 ) q . add ( stepNumB ); // If lastDigit is 9 then only possible // digit after 9 can be 8 for a Stepping // Number else if ( lastDigit == 9 ) q . add ( stepNumA ); else { q . add ( stepNumA ); q . add ( stepNumB ); } } } // Prints all stepping numbers in range [n m] // using BFS. public static void displaySteppingNumbers ( int n int m ) { // For every single digit Number 'i' // find all the Stepping Numbers // starting with i for ( int i = 0 ; i <= 9 ; i ++ ) bfs ( n m i ); } //Driver code public static void main ( String args [] ) { int n = 0 m = 21 ; // Display Stepping Numbers in // the range [nm] displaySteppingNumbers ( n m ); } }
Python3 # A Python3 program to find all the Stepping Number from N=n # to m using BFS Approach # Prints all stepping numbers reachable from num # and in range [n m] def bfs ( n m num ) : # Queue will contain all the stepping Numbers q = [] q . append ( num ) while len ( q ) > 0 : # Get the front element and pop from the queue stepNum = q [ 0 ] q . pop ( 0 ); # If the Stepping Number is in the range # [n m] then display if ( stepNum <= m and stepNum >= n ) : print ( stepNum end = ' ' ) # If Stepping Number is 0 or greater than m # no need to explore the neighbors if ( num == 0 or stepNum > m ) : continue # Get the last digit of the currently visited # Stepping Number lastDigit = stepNum % 10 # There can be 2 cases either digit to be # appended is lastDigit + 1 or lastDigit - 1 stepNumA = stepNum * 10 + ( lastDigit - 1 ) stepNumB = stepNum * 10 + ( lastDigit + 1 ) # If lastDigit is 0 then only possible digit # after 0 can be 1 for a Stepping Number if ( lastDigit == 0 ) : q . append ( stepNumB ) #If lastDigit is 9 then only possible #digit after 9 can be 8 for a Stepping #Number elif ( lastDigit == 9 ) : q . append ( stepNumA ) else : q . append ( stepNumA ) q . append ( stepNumB ) # Prints all stepping numbers in range [n m] # using BFS. def displaySteppingNumbers ( n m ) : # For every single digit Number 'i' # find all the Stepping Numbers # starting with i for i in range ( 10 ) : bfs ( n m i ) # Driver code n m = 0 21 # Display Stepping Numbers in the # range [nm] displaySteppingNumbers ( n m ) # This code is contributed by divyeshrabadiya07.
C# // A C# program to find all the Stepping Number in // range [n m] using System ; using System.Collections.Generic ; public class GFG { // Prints all stepping numbers reachable from num // and in range [n m] static void bfs ( int n int m int num ) { // Queue will contain all the stepping Numbers Queue < int > q = new Queue < int > (); q . Enqueue ( num ); while ( q . Count != 0 ) { // Get the front element and pop from // the queue int stepNum = q . Dequeue (); // If the Stepping Number is in // the range [nm] then display if ( stepNum <= m && stepNum >= n ) { Console . Write ( stepNum + ' ' ); } // If Stepping Number is 0 or greater // then m no need to explore the neighbors if ( stepNum == 0 || stepNum > m ) continue ; // Get the last digit of the currently // visited Stepping Number int lastDigit = stepNum % 10 ; // There can be 2 cases either digit // to be appended is lastDigit + 1 or // lastDigit - 1 int stepNumA = stepNum * 10 + ( lastDigit - 1 ); int stepNumB = stepNum * 10 + ( lastDigit + 1 ); // If lastDigit is 0 then only possible // digit after 0 can be 1 for a Stepping // Number if ( lastDigit == 0 ) q . Enqueue ( stepNumB ); // If lastDigit is 9 then only possible // digit after 9 can be 8 for a Stepping // Number else if ( lastDigit == 9 ) q . Enqueue ( stepNumA ); else { q . Enqueue ( stepNumA ); q . Enqueue ( stepNumB ); } } } // Prints all stepping numbers in range [n m] // using BFS. static void displaySteppingNumbers ( int n int m ) { // For every single digit Number 'i' // find all the Stepping Numbers // starting with i for ( int i = 0 ; i <= 9 ; i ++ ) bfs ( n m i ); } // Driver code static public void Main () { int n = 0 m = 21 ; // Display Stepping Numbers in // the range [nm] displaySteppingNumbers ( n m ); } } // This code is contributed by avanitrachhadiya2155
JavaScript < script > // A Javascript program to find all // the Stepping Number in // range [n m] // Prints all stepping numbers // reachable from num // and in range [n m] function bfs ( n m num ) { // Queue will contain all the // stepping Numbers let q = []; q . push ( num ); while ( q . length != 0 ) { // Get the front element and pop from // the queue let stepNum = q . shift (); // If the Stepping Number is in // the range [nm] then display if ( stepNum <= m && stepNum >= n ) { document . write ( stepNum + ' ' ); } // If Stepping Number is 0 or greater // then m no need to explore the neighbors if ( stepNum == 0 || stepNum > m ) continue ; // Get the last digit of the currently // visited Stepping Number let lastDigit = stepNum % 10 ; // There can be 2 cases either digit // to be appended is lastDigit + 1 or // lastDigit - 1 let stepNumA = stepNum * 10 + ( lastDigit - 1 ); let stepNumB = stepNum * 10 + ( lastDigit + 1 ); // If lastDigit is 0 then only possible // digit after 0 can be 1 for a Stepping // Number if ( lastDigit == 0 ) q . push ( stepNumB ); // If lastDigit is 9 then only possible // digit after 9 can be 8 for a Stepping // Number else if ( lastDigit == 9 ) q . push ( stepNumA ); else { q . push ( stepNumA ); q . push ( stepNumB ); } } } // Prints all stepping numbers in range [n m] // using BFS. function displaySteppingNumbers ( n m ) { // For every single digit Number 'i' // find all the Stepping Numbers // starting with i for ( let i = 0 ; i <= 9 ; i ++ ) bfs ( n m i ); } // Driver code let n = 0 m = 21 ; // Display Stepping Numbers in // the range [nm] displaySteppingNumbers ( n m ); // This code is contributed by unknown2108 < /script>
Uitvoer
0 1 10 12 2 21 3 4 5 6 7 8 9
Op DFS gebaseerde oplossing:
C++ // A C++ program to find all the Stepping Numbers // in range [n m] using DFS Approach #include using namespace std ; // Prints all stepping numbers reachable from num // and in range [n m] void dfs ( int n int m int stepNum ) { // If Stepping Number is in the // range [nm] then display if ( stepNum <= m && stepNum >= n ) cout < < stepNum < < ' ' ; // If Stepping Number is 0 or greater // than m then return if ( stepNum == 0 || stepNum > m ) return ; // Get the last digit of the currently // visited Stepping Number int lastDigit = stepNum % 10 ; // There can be 2 cases either digit // to be appended is lastDigit + 1 or // lastDigit - 1 int stepNumA = stepNum * 10 + ( lastDigit -1 ); int stepNumB = stepNum * 10 + ( lastDigit + 1 ); // If lastDigit is 0 then only possible // digit after 0 can be 1 for a Stepping // Number if ( lastDigit == 0 ) dfs ( n m stepNumB ); // If lastDigit is 9 then only possible // digit after 9 can be 8 for a Stepping // Number else if ( lastDigit == 9 ) dfs ( n m stepNumA ); else { dfs ( n m stepNumA ); dfs ( n m stepNumB ); } } // Method displays all the stepping // numbers in range [n m] void displaySteppingNumbers ( int n int m ) { // For every single digit Number 'i' // find all the Stepping Numbers // starting with i for ( int i = 0 ; i <= 9 ; i ++ ) dfs ( n m i ); } //Driver program to test above function int main () { int n = 0 m = 21 ; // Display Stepping Numbers in // the range [nm] displaySteppingNumbers ( n m ); return 0 ; }
Java // A Java program to find all the Stepping Numbers // in range [n m] using DFS Approach import java.util.* ; class Main { // Method display's all the stepping numbers // in range [n m] public static void dfs ( int n int m int stepNum ) { // If Stepping Number is in the // range [nm] then display if ( stepNum <= m && stepNum >= n ) System . out . print ( stepNum + ' ' ); // If Stepping Number is 0 or greater // than m then return if ( stepNum == 0 || stepNum > m ) return ; // Get the last digit of the currently // visited Stepping Number int lastDigit = stepNum % 10 ; // There can be 2 cases either digit // to be appended is lastDigit + 1 or // lastDigit - 1 int stepNumA = stepNum * 10 + ( lastDigit - 1 ); int stepNumB = stepNum * 10 + ( lastDigit + 1 ); // If lastDigit is 0 then only possible // digit after 0 can be 1 for a Stepping // Number if ( lastDigit == 0 ) dfs ( n m stepNumB ); // If lastDigit is 9 then only possible // digit after 9 can be 8 for a Stepping // Number else if ( lastDigit == 9 ) dfs ( n m stepNumA ); else { dfs ( n m stepNumA ); dfs ( n m stepNumB ); } } // Prints all stepping numbers in range [n m] // using DFS. public static void displaySteppingNumbers ( int n int m ) { // For every single digit Number 'i' // find all the Stepping Numbers // starting with i for ( int i = 0 ; i <= 9 ; i ++ ) dfs ( n m i ); } // Driver code public static void main ( String args [] ) { int n = 0 m = 21 ; // Display Stepping Numbers in // the range [nm] displaySteppingNumbers ( n m ); } }
Python3 # A Python3 program to find all the Stepping Numbers # in range [n m] using DFS Approach # Prints all stepping numbers reachable from num # and in range [n m] def dfs ( n m stepNum ) : # If Stepping Number is in the # range [nm] then display if ( stepNum <= m and stepNum >= n ) : print ( stepNum end = ' ' ) # If Stepping Number is 0 or greater # than m then return if ( stepNum == 0 or stepNum > m ) : return # Get the last digit of the currently # visited Stepping Number lastDigit = stepNum % 10 # There can be 2 cases either digit # to be appended is lastDigit + 1 or # lastDigit - 1 stepNumA = stepNum * 10 + ( lastDigit - 1 ) stepNumB = stepNum * 10 + ( lastDigit + 1 ) # If lastDigit is 0 then only possible # digit after 0 can be 1 for a Stepping # Number if ( lastDigit == 0 ) : dfs ( n m stepNumB ) # If lastDigit is 9 then only possible # digit after 9 can be 8 for a Stepping # Number elif ( lastDigit == 9 ) : dfs ( n m stepNumA ) else : dfs ( n m stepNumA ) dfs ( n m stepNumB ) # Method displays all the stepping # numbers in range [n m] def displaySteppingNumbers ( n m ) : # For every single digit Number 'i' # find all the Stepping Numbers # starting with i for i in range ( 10 ) : dfs ( n m i ) n m = 0 21 # Display Stepping Numbers in # the range [nm] displaySteppingNumbers ( n m ) # This code is contributed by divyesh072019.
C# // A C# program to find all the Stepping Numbers // in range [n m] using DFS Approach using System ; public class GFG { // Method display's all the stepping numbers // in range [n m] static void dfs ( int n int m int stepNum ) { // If Stepping Number is in the // range [nm] then display if ( stepNum <= m && stepNum >= n ) Console . Write ( stepNum + ' ' ); // If Stepping Number is 0 or greater // than m then return if ( stepNum == 0 || stepNum > m ) return ; // Get the last digit of the currently // visited Stepping Number int lastDigit = stepNum % 10 ; // There can be 2 cases either digit // to be appended is lastDigit + 1 or // lastDigit - 1 int stepNumA = stepNum * 10 + ( lastDigit - 1 ); int stepNumB = stepNum * 10 + ( lastDigit + 1 ); // If lastDigit is 0 then only possible // digit after 0 can be 1 for a Stepping // Number if ( lastDigit == 0 ) dfs ( n m stepNumB ); // If lastDigit is 9 then only possible // digit after 9 can be 8 for a Stepping // Number else if ( lastDigit == 9 ) dfs ( n m stepNumA ); else { dfs ( n m stepNumA ); dfs ( n m stepNumB ); } } // Prints all stepping numbers in range [n m] // using DFS. public static void displaySteppingNumbers ( int n int m ) { // For every single digit Number 'i' // find all the Stepping Numbers // starting with i for ( int i = 0 ; i <= 9 ; i ++ ) dfs ( n m i ); } // Driver code static public void Main () { int n = 0 m = 21 ; // Display Stepping Numbers in // the range [nm] displaySteppingNumbers ( n m ); } } // This code is contributed by rag2127.
JavaScript < script > // A Javascript program to find all the Stepping Numbers // in range [n m] using DFS Approach // Method display's all the stepping numbers // in range [n m] function dfs ( n m stepNum ) { // If Stepping Number is in the // range [nm] then display if ( stepNum <= m && stepNum >= n ) document . write ( stepNum + ' ' ); // If Stepping Number is 0 or greater // than m then return if ( stepNum == 0 || stepNum > m ) return ; // Get the last digit of the currently // visited Stepping Number let lastDigit = stepNum % 10 ; // There can be 2 cases either digit // to be appended is lastDigit + 1 or // lastDigit - 1 let stepNumA = stepNum * 10 + ( lastDigit - 1 ); let stepNumB = stepNum * 10 + ( lastDigit + 1 ); // If lastDigit is 0 then only possible // digit after 0 can be 1 for a Stepping // Number if ( lastDigit == 0 ) dfs ( n m stepNumB ); // If lastDigit is 9 then only possible // digit after 9 can be 8 for a Stepping // Number else if ( lastDigit == 9 ) dfs ( n m stepNumA ); else { dfs ( n m stepNumA ); dfs ( n m stepNumB ); } } // Prints all stepping numbers in range [n m] // using DFS. function displaySteppingNumbers ( n m ) { // For every single digit Number 'i' // find all the Stepping Numbers // starting with i for ( let i = 0 ; i <= 9 ; i ++ ) dfs ( n m i ); } // Driver code let n = 0 m = 21 ; // Display Stepping Numbers in // the range [nm] displaySteppingNumbers ( n m ); // This code is contributed by ab2127 < /script>
Uitvoer
0 1 10 12 2 21 3 4 5 6 7 8 9
Tijdcomplexiteit:O(N log N)
Ruimtecomplexiteit: O (N) hier is N het aantal stapnummers binnen het bereik.
Dit Vind Je Misschien Leuk
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