ステップ番号
GfG Practice で試してみる 
#practiceLinkDiv { 表示: なし !重要; }
出力
#practiceLinkDiv { 表示: なし !重要; } 2 つの整数「n」と「m」を指定すると、範囲 [n m] 内のすべてのステップ番号が検索されます。番号が呼ばれます ステップ数 すべての隣接する数字の差の絶対値が 1 である場合。321 はステッピング番号ですが、421 はステッピング番号ではありません。
例:
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 絶対差が 1 つある数値 試してみてください!
方法 1: ブルート フォース アプローチ
この方法では、総当たりのアプローチを使用して、n から m までのすべての整数を反復処理し、それがステップ数であるかどうかを確認します。
// 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>
出力
0 1 2 3 4 5 6 7 8 9 10 12 21
方法 2: BFS/DFS を使用する
アイデアは、 幅優先検索 / 深さ優先検索 横断。
グラフを作成するにはどうすればよいですか?
グラフ内のすべてのノードはステップ番号を表します。 V を U から変換できる場合、ノード U から V への有向エッジが存在します。(U と V はステッピング数です) ステッピング数 V は、次の方法で U から変換できます。
最後の桁 U の最後の桁を指します (つまり、U % 10)
隣接する番号 V 可能性があるのは次のとおりです:
- U*10 + lastDigit + 1 (隣接 A)
- U*10 + lastDigit – 1 (隣接 B)
上記の操作を適用すると、新しい数字が U に追加されます。これは、lastDigit-1 または lastDigit+1 のいずれかになるため、U から形成される新しい数値 V もステッピング番号になります。
したがって、すべてのノードには最大 2 つの隣接ノードがあります。
エッジケース: Uの下一桁が または 9
- すべての 1 桁の数値はステップ番号とみなされ、すべての桁に対する bfs のトラバーサルにより、その桁から始まるすべてのステップ番号が得られます。
- [09] のすべての番号に対して bfs/dfs トラバーサルを実行します。
ソース/開始ノードは何になりますか?
注記： ノード 0 の場合、01 012 010 につながり、ノード 1 から始まる BFS トラバーサルでカバーされるため、BFS トラバーサル中に近隣ノードを探索する必要はありません。
0から21までのすべてのステップ番号を検索する例
-> 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.
BFS ベースのソリューション:
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>
出力
0 1 10 12 2 21 3 4 5 6 7 8 9
DFS ベースのソリューション:
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>
出力
0 1 10 12 2 21 3 4 5 6 7 8 9
時間計算量:O(N log N)
空間の複雑さ:O(N) ここで、N は範囲内のステップ数の数です。
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