El problema de la subsecuencia creciente más larga (LIS) es encontrar la longitud de la subsecuencia más larga de una secuencia dada de modo que todos los elementos de la subsecuencia se clasifiquen en orden creciente. Por ejemplo, la longitud de LIS para {10, 22, 9, 33, 21, 50, 41, 60, 80} es 6 y LIS es {10, 22, 33, 50, 60, 80}.
Más ejemplos:
Input : arr[] = {3, 10, 2, 1, 20} Output : Length of LIS = 3 The longest increasing subsequence is 3, 10, 20 Input : arr[] = {3, 2} Output : Length of LIS = 1 The longest increasing subsequences are {3} and {2} Input : arr[] = {50, 3, 10, 7, 40, 80} Output : Length of LIS = 4 The longest increasing subsequence is {3, 7, 40, 80}
Subestructura óptima:
Sea arr[0..n-1] la array de entrada y L(i) la longitud del LIS que termina en el índice i, de modo que arr[i] es el último elemento del LIS.
Entonces, L(i) puede escribirse recursivamente como:
L(i) = 1 + max( L(j) ) donde 0 < j < i y arr[j] < arr[i]; o
L(i) = 1, si tal j no existe. +
Para encontrar el LIS para una array dada, necesitamos devolver max(L(i)) donde 0 < i < n.
Por lo tanto, vemos que el problema LIS satisface la propiedad de subestructura óptima ya que el problema principal se puede resolver utilizando soluciones a los subproblemas.
A continuación se muestra una implementación recursiva simple del problema LIS. Sigue la estructura recursiva discutida anteriormente.
C++
/* A Naive C++ recursive implementation of LIS problem */ #include <iostream> using namespace std; /* To make use of recursive calls, this function must return two things: 1) Length of LIS ending with element arr[n-1]. We use max_ending_here for this purpose 2) Overall maximum as the LIS may end with an element before arr[n-1] max_ref is used this purpose. The value of LIS of full array of size n is stored in *max_ref which is our final result */ int _lis(int arr[], int n, int* max_ref) { /* Base case */ if (n == 1) return 1; // 'max_ending_here' is length of LIS ending with arr[n-1] int res, max_ending_here = 1; /* Recursively get all LIS ending with arr[0], arr[1] ... arr[n-2]. If arr[i-1] is smaller than arr[n-1], and max ending with arr[n-1] needs to be updated, then update it */ for(int i = 1; i < n; i++) { res = _lis(arr, i, max_ref); if (arr[i - 1] < arr[n - 1] && res + 1 > max_ending_here) max_ending_here = res + 1; } // Compare max_ending_here with the overall max. And // update the overall max if needed if (*max_ref < max_ending_here) *max_ref = max_ending_here; // Return length of LIS ending with arr[n-1] return max_ending_here; } // The wrapper function for _lis() int lis(int arr[], int n) { // The max variable holds the result int max = 1; // The function _lis() stores its result in max _lis(arr, n, &max); // returns max return max; } // Driver code int main() { int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "Length of lis is " << lis(arr, n) << "\n"; return 0; } // This code is contributed by Shubhamsingh10
C
/* A Naive C recursive implementation of LIS problem */ #include <stdio.h> #include <stdlib.h> /* To make use of recursive calls, this function must return two things: 1) Length of LIS ending with element arr[n-1]. We use max_ending_here for this purpose 2) Overall maximum as the LIS may end with an element before arr[n-1] max_ref is used this purpose. The value of LIS of full array of size n is stored in *max_ref which is our final result */ int _lis(int arr[], int n, int* max_ref) { /* Base case */ if (n == 1) return 1; // 'max_ending_here' is length of LIS ending with arr[n-1] int res, max_ending_here = 1; /* Recursively get all LIS ending with arr[0], arr[1] ... arr[n-2]. If arr[i-1] is smaller than arr[n-1], and max ending with arr[n-1] needs to be updated, then update it */ for (int i = 1; i < n; i++) { res = _lis(arr, i, max_ref); if (arr[i - 1] < arr[n - 1] && res + 1 > max_ending_here) max_ending_here = res + 1; } // Compare max_ending_here with the overall max. And // update the overall max if needed if (*max_ref < max_ending_here) *max_ref = max_ending_here; // Return length of LIS ending with arr[n-1] return max_ending_here; } // The wrapper function for _lis() int lis(int arr[], int n) { // The max variable holds the result int max = 1; // The function _lis() stores its result in max _lis(arr, n, &max); // returns max return max; } /* Driver program to test above function */ int main() { int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 }; int n = sizeof(arr) / sizeof(arr[0]); printf("Length of lis is %d\n", lis(arr, n)); return 0; }
Length of lis is 5
Complejidad de tiempo : O(n)
Espacio Auxiliar : O(1)
Consulte el artículo completo sobre programación dinámica | ¡Establezca 3 (Subsecuencia creciente más larga) para obtener más detalles!
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA