Dado o número de dígitos n em um número, imprima todos os números de n dígitos cujos dígitos aumentam estritamente da esquerda para a direita.
Exemplos:
Input: n = 2 Output: 01 02 03 04 05 06 07 08 09 12 13 14 15 16 17 18 19 23 24 25 26 27 28 29 34 35 36 37 38 39 45 46 47 48 49 56 57 58 59 67 68 69 78 79 89 Input: n = 3 Output: 012 013 014 015 016 017 018 019 023 024 025 026 027 028 029 034 035 036 037 038 039 045 046 047 048 049 056 057 058 059 067 068 069 078 079 089 123 124 125 126 127 128 129 134 135 136 137 138 139 145 146 147 148 149 156 157 158 159 167 168 169 178 179 189 234 235 236 237 238 239 245 246 247 248 249 256 257 258 259 267 268 269 278 279 289 345 346 347 348 349 356 357 358 359 367 368 369 378 379 389 456 457 458 459 467 468 469 478 479 489 567 568 569 578 579 589 678 679 689 789 Input: n = 1 Output: 0 1 2 3 4 5 6 7 8 9
A ideia é usar recursão. Começamos na posição mais à esquerda de um possível número de N dígitos e o preenchemos a partir do conjunto de todos os dígitos maiores que o dígito anterior. ou seja, preencha a posição atual com dígitos (i a 9] onde i é o dígito anterior. Depois de preencher a posição atual, recorremos à próxima posição com números estritamente crescentes.
Abaixo está a implementação da ideia acima –
// C++ program to print all n-digit numbers whose digits // are strictly increasing from left to right #include
// Java program to print all n-digit numbers whose digits // are strictly increasing from left to right import java.io.*; class Increasing { // Function to print all n-digit numbers whose digits // are strictly increasing from left to right. // out --> Stores current output number as string // start --> Current starting digit to be considered void findStrictlyIncreasingNum(int start String out int n) { // If number becomes N-digit print it if (n == 0) { System.out.print(out + ' '); return; } // start from (prev digit + 1) till 9 for (int i = start; i <= 9; i++) { // append current digit to number String str = out + Integer.toString(i); // recurse for next digit findStrictlyIncreasingNum(i + 1 str n - 1); } } // Driver code for above function public static void main(String args[])throws IOException { Increasing obj = new Increasing(); int n = 3; obj.findStrictlyIncreasingNum(0 ' ' n); } }
# Python3 program to print all n-digit numbers # whose digits are strictly increasing # from left to right # Function to print all n-digit numbers # whose digits are strictly increasing # from left to right. # out --> Stores current output # number as string # start --> Current starting digit # to be considered def findStrictlyIncreasingNum(start out n): # If number becomes N-digit print if (n == 0): print(out end = ' ') return # start from (prev digit + 1) till 9 for i in range(start 10): # append current digit to number str1 = out + str(i) # recurse for next digit findStrictlyIncreasingNum(i + 1 str1 n - 1) # Driver code n = 3 findStrictlyIncreasingNum(0 '' n) # This code is contributed by Mohit Kumar
// C# program to print all n-digit numbers // whose digits are strictly increasing // from left to right using System; class GFG { // Function to print all n-digit numbers // whose digits are strictly increasing // from left to right. out --> Stores // current output number as string // start --> Current starting digit to // be considered static void findStrictlyIncreasingNum(int start string Out int n) { // If number becomes N-digit print it if (n == 0) { Console.Write(Out + ' '); return; } // start from (prev digit + 1) till 9 for (int i = start; i <= 9; i++) { // append current digit to number string str = Out + Convert.ToInt32(i); // recurse for next digit findStrictlyIncreasingNum(i + 1 str n - 1); } } // Driver code for above function public static void Main() { int n = 3; findStrictlyIncreasingNum(0 ' ' n); } } // This code is contributed by Sam007.
<script> // Javascript program to print all // n-digit numbers whose digits // are strictly increasing from // left to right // Function to print all // n-digit numbers whose digits // are strictly increasing // from left to right. // out --> Stores current // output number as string // start --> Current starting // digit to be considered function findStrictlyIncreasingNum(startoutn) { // If number becomes N-digit print it if (n == 0) { document.write(out + ' '); return; } // start from (prev digit + 1) till 9 for (let i = start; i <= 9; i++) { // append current digit to number let str = out + i.toString(); // recurse for next digit findStrictlyIncreasingNum(i + 1 str n - 1); } } // Driver code for above function let n = 3; findStrictlyIncreasingNum(0 ' ' n); // This code is contributed by unknown2108 </script>
Saída:
012 013 014 015 016 017 018 019 023 024 025 026 027 028 029 034 035 036 037 038 039 045 046 047 048 049 056 057 058 059 067 068 069 078 079 089 123 124 125 126 127 128 129 134 135 136 137 138 139 145 146 147 148 149 156 157 158 159 167 168 169 178 179 189 234 235 236 237 238 239 245 246 247 248 249 256 257 258 259 267 268 269 278 279 289 345 346 347 348 349 356 357 358 359 367 368 369 378 379 389 456 457 458 459 467 468 469 478 479 489 567 568 569 578 579 589 678 679 689 789
Complexidade de tempo: SOBRE!)
Espaço Auxiliar: SOBRE)
Exercício: Imprima todos os números de n dígitos cujos dígitos diminuem estritamente da esquerda para a direita.