숫자의 자릿수 n이 주어지면 자릿수가 왼쪽에서 오른쪽으로 엄격하게 증가하는 모든 n자리 숫자를 인쇄합니다.
예:
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
내 모니터 화면 크기는 몇이야
아이디어는 재귀를 사용하는 것입니다. 가능한 N 자리 숫자의 가장 왼쪽 위치에서 시작하여 이전 숫자보다 큰 모든 숫자 집합에서 채웁니다. 즉, 현재 위치를 숫자(i ~ 9]로 채웁니다. 여기서 i는 이전 숫자입니다. 현재 위치를 채운 후 엄격하게 증가하는 숫자를 사용하여 다음 위치로 반복합니다.
아래는 위의 아이디어를 구현한 것입니다.
// 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>
산출:
클래스가 여러 클래스를 확장할 수 있나요?
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
시간 복잡도: 에!)
보조 공간: 에)
운동: 자릿수가 왼쪽에서 오른쪽으로 엄격하게 감소하는 모든 n자리 숫자를 인쇄합니다.