Write a recursive function in java to find the maximum value of an array

How to use java to achieve the maximum value of an array recursively

public static void main(String[] rags){

int [] aim = new int[100];

int point = 0;

//....Initialize the array here

int max = aim[0];

max = getMax(max,point,aim);

//...Other processing

}

//Recursive method

public int getMax(int ​​max,int point,int[] aim){

if(point==aim.length) //Critical value

return max;

//When the critical value is not reached, take the max value and perform recursion

max = max >= aim[point] ? max : aim[point];

return getMax(max,point 1,aim);

}

How to implement the recursive algorithm of binary search in java

public class Binary recursive search {

public static void main(String[] args) {

//Define the array. Note that the binary search array must be an ordered array!

int[] arr = { 1, 3, 5, 7, 9, 11, 13, 15, 17 };

//Accept the return value after the search: index value, if not, it is -1;

//Test search element: 9

int a=binary(arr, 9, 0, arr.length - 1);

System.out.println("The index position of the number being searched is:" a);

}

//The parameter list is: the array to be searched, the number to search for, the head index, and the tail index!

public static int binary(int[] arr, int key, int star, int end)//recursion

{

//Create every time you come in, the intermediate index value!

int mid = (star end) / 2;

//If the number being searched is less than the head or tail, or the head index is greater than the tail index, it means there is no such number and -1 is returned;

if (key arr[end] || star > end) {

return -1;

}

//If the middle value is less than the number being searched, redefine the header index and move it to the middle 1 position, filtering out half of the numbers!

if (arr[mid]

//Start recursion!

return binary(arr, key, mid 1, end);

//Otherwise, if the middle value is greater than the number being searched, the tail index will be moved to the middle -1 position and half of the numbers will be filtered out!

} else if (arr[mid] > key) {

//Start recursion!

return binary(arr,key, star, mid - 1);

} else {

//If not, it is found and returns to the index!

return mid;

}

}

}

How is Java's recursion executed and how is the order executed?

factest(8) enters the factest function, if(n==1) return 1; // If not established, execute else else return n*factest(n-1); // The return value is 8*factest(7)

factest(7) enters the factest function, if(n==1) return 1; // If not established, execute else

else return n*factest(n-1); // The return value is 7*factest(6)

……

Until N=1, at this time if(n==1) return 1; // Established, the return value is 1, that is, 1!=1

Then calculate the return value of factest(2) as: 2*factest(1) = 2

Then continue to calculate the return value of factest(3): 3*factest(2) = 6

...... Until N=8, get factest(8) = 8*factest(7) = 40320

How to use recursion to solve this problem in JAVA? Master

The Java recursive program you want to write is as follows:

import java.util.Scanner;

public class GGG {

public static void main(String[] args) {

int N = 0;

Scanner sc=new Scanner(System.in);

int num=sc.nextInt();

for(int n=0;n

N=sc.nextInt();

int a[]=new int[N];

for(int i=0;i

a[i]=sc.nextInt();

}

System.out.print("case" (n 1) ":");

process(a,0);

System.out.println();

}

}

private static void process(int[] a, int n) {

if(n==0){

if(isPrime(a[n 1]))

System.out.print(1 ＂ ＂);

else

System.out.print(0 ＂ ＂);

}else if(n==a.length-1){

if(isPrime(a[n-1]))

System.out.print(1 ＂ ＂);

else

System.out.print(0 ＂ ＂);

return;

}else{

if(isPrime(a[n-1])&isPrime(a[n 1]))

System.out.print(2 ＂ ＂);

else if(isPrime(a[n-1])||isPrime(a[n 1]))

System.out.print(1 ＂ ＂);

else

System.out.print(0 ＂ ＂);

}

process(a,n 1);

}

public static boolean isPrime(int num) {

int i;

for(i=2;i

if(num%i==0)

break;

}

if(i==num){

return true;

}

return false;

}

}operation result:

2

5

5 7 2 9 13

case 1:1 2 1 2 0

3

10 4 5

case 2:0 1 0

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