分治JAVA向量
这是我的代码: 导入java.io.IOException; 导入java.util.Scanner分治JAVA向量,java,algorithm,divide,divide-and-conquer,Java,Algorithm,Divide,Divide And Conquer,这是我的代码: 导入java.io.IOException; 导入java.util.Scanner public class Main { static int operaciones; public static int GetFrequency(int A[],int valor, int izq, int der){ int cont = 0; for(int i=izq; i<= der; i++){ if(
public class Main {
static int operaciones;
public static int GetFrequency(int A[],int valor, int izq, int der){
int cont = 0;
for(int i=izq; i<= der; i++){
if(A[i] == valor){
cont++;
}
operaciones++;
}
return cont;
}
public static void print(int A[]){
System.out.println("El vector Ingresado Fue: ");
System.out.println("--------------------------");
for(int i=1; i<A.length;i++){
System.out.print(A[i] + "\n");
}
}
public static void ingresar(int A[], int elementos, Scanner sc){
System.out.println("Ingrese los elementos");
for(int i=1; i<A.length;i++){
elementos = sc.nextInt();
A[i]=elementos;
}
}
public static boolean mayoritario(int A[],int i, int j){
if(i<=j){
int valor = 0;
for(int x = i; i <j; x++){
valor = GetFrequency(A, A[x], i, j);
operaciones++;
}
if(((A.length-1)%2) == 0){
if(valor > (A.length-1)/2){
return true;
}
}else{
if(valor >= (A.length/2)){
return true;
}
}
return false;
}
if(mayoritario(A,i,j/2)){
return true;
}if(mayoritario(A,((i+(j/2))),j)){
return true;
}
return false;
}
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
System.out.println("Ingrese cantidad de elementos");
int n = sc.nextInt();
int A [] = new int [n+1];
int elementos = 0;
int maximo = n;
int minimo = 1;
ingresar(A,elementos,sc);
print(A);
System.out.println("-------------------------");
if(mayoritario(A,minimo,maximo)){
System.out.println("Existe un mayoritario");
}else{
System.out.println("No existe un mayoritario");
}
System.out.println("Operaciones = "+operaciones);
}
}
公共类主{
静态内部操作;
公共静态int GetFrequency(int A[],int valor,int izq,int der){
int cont=0;
对于(int i=izq;i而不是在每次迭代时运行GetFrequency
,将问题分为两个步骤。在步骤1中,将频率列表成一个映射。您应该能够在O(n)中完成此操作。在步骤2中,迭代映射并找到与最大值相关的键