尝试使用递归

Question

获得偶数的总和在阵列我想获得偶数的总和的数组：

public static void main(String[] args) { 
    int[] array = new int[4]; 
    array[0] = 1; 
    array[1] = 2; 
    array[2] = 2; 
    array[3] = 4; 

    System.out.println("Count even: " + countE(array, 0));  
} 

public static int countE(int[] arr, int head) { 
     if (arr.length == head) { 
     return -1; 
     } else if (arr[head] % 2 == 0) {    

      return 1 + countE(arr, head + 1); 
    } else { 

     return 0 + countE(arr, head + 1); 
    } 
} 

Answer 1

if (arr.length == head) { 
     return -1; 

是给你的问题的兄弟......你减去1从最终的个性化......尝试做

if (arr.length == head) { 
     return 0; 

Answer 2

它可以通过递归来完成，要添加的数字错误

 else if (arr[head]%2==0) 
     { 
      return arr[head] + countHead(arr, head + 1); 
     } 

Answer 3

它给你2，因为在上一次递归时，它会扣除1 - 你的基本情况是错误的。反而返回0。

Answer 4

我希望我不会为你透露太多，它带走了学习的兴奋！

public static int countE(int[] array, int index) { 
    int currentNumber = array[index]; 
    if(index == array.length) { 
    // terminate it by returning a value that does not affect the sum (what integer is this?) 
    } else { 
    // check if even, if even, add the current number, else, add zero (both are recursive calls) 
    if(isEven(currentNumber)) { 
     return // your recursive call if even, currentNumber should be added 
    } else { 
     return // your recursive call if not, currentNumber should not be added 
    } 
    } 
} 

public static boolean isEven(int number) { 
    return (number % 2 == 0) ? true : false; 
} 

Answer 5

Hlo。好友..试试这个

package com.smk.recursion; 

public class SumOfEvenNoFromArray { 
public static void main(String[] args) { 

    int[] a = new int[5]; 
    a[0] = 2; 
    a[1] = 5; 
    a[2] = 4; 
    a[3] = 4; 
    a[4] = 4; 

    System.out.println("Sum of Odd Numbers : "+sumEvenNoRecursion(a,(a.length-1))); 

} 

public static int sumEvenNoRecursion(int[] arr, int index){ 
    int sum; 
    if(index == -1){ 
     return 0; 
    }else if(arr[index]%2 == 0){   
     sum = arr[index] + sumEvenNoRecursion(arr, --index);    
    }else{ 
     sum = sumEvenNoRecursion(arr, --index);    
    } 
    return sum; 


} 
} 

//输出

总和奇数：14

Answer 6

你可以使用一个辅助功能，为您的递归。

public static int sumofEvens(int[] list){ 
    int sum = sumHelper(list, 0, 0); 
    return sum; 
} 
private static int sumHelper(int[] list, int i, int sum){ 
    if(i < list.length){ 
     if(list[i] % 2 == 0){ 
      sum += list[i]; 
     } 
     sum = sumHelper(list, i+1, sum); 
    } 
    return sum; 
}