Question

关于Java的recursion的问题

想要一个程序。要求你输入数字。输入0的时候输出以下
The minimum number is
The sum of positive numbers is
The sum of numbers at even indexes is
The total count of odd integers is

要求不能用loop在
public static int findMin(int[] numbers, int startIndex, int endIndex)
public static int computePositiveSum(int[] numbers, int startIndex, int endIndex)
public static int computeSumAtEven(int[] numbers, int startIndex, int endIndex)
public static int countOdd(int[] numbers, int startIndex, int endIndex)

但是输入array的时候可以用loop饿
各位大神求助啦。
小弟初学是在不太会这个

Answer 1

全当练习，自己不做，总学不会，用的分治算法

import java.util.Scanner;
public class Test{
public static int findMin(int[] numbers, int startIndex, int endIndex){
if(startIndex>=endIndex-1)
return numbers[startIndex];
int mid=(startIndex+endIndex)/2;
//System.out.println("start:"+startIndex+"end:"+endIndex+" mid:"+mid);
int left=findMin(numbers, startIndex, mid);
int right=findMin(numbers, mid, endIndex);
return left<=right?left:right;

}
public static int computePositiveSum(int[] numbers, int startIndex, int endIndex){
if(startIndex>=endIndex-1)
return numbers[startIndex]>0?numbers[startIndex]:0;
int mid=(startIndex+endIndex)/2;
return computePositiveSum(numbers, startIndex, mid)+computePositiveSum(numbers, mid, endIndex);
}
//由于java数组的下标是从0算起，0,1,2,3,4,5，取偶数0,2,4
public static int computeSumAtEven(int[] numbers, int startIndex, int endIndex){
if(startIndex>=endIndex-1)
return startIndex%2==0?numbers[startIndex]:0;
int mid=(startIndex+endIndex)/2;
return computeSumAtEven(numbers, startIndex, mid)+computeSumAtEven(numbers, mid, endIndex);
}
public static int countOdd(int[] numbers, int startIndex, int endIndex){
if(startIndex>=endIndex-1)
return numbers[startIndex]%2==0?0:1;
int mid=(startIndex+endIndex)/2;
return countOdd(numbers, startIndex, mid)+countOdd(numbers, mid, endIndex);
}
public static void main(String[] args) {

Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
//int a[]={3,4,1,6,7};
System.out.println("The minimum number is "+findMin(a,0,a.length)+
"\nThe sum of positive numbers is "+computePositiveSum(a,0,a.length)+
"\nThe sum of numbers at even indexes is "+computeSumAtEven(a,0,a.length)+
"\nThe total count of odd integers is "+countOdd(a,0,a.length)
);
}
}

5
4 3 1 5 -2
The minimum number is -2
The sum of positive numbers is 13
The sum of numbers at even indexes is 3
The total count of odd integers is 3