blocks|key|1622473|text|问题是a[n-1]是一个int，而sumOfArray需要一个int数组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1622474|提示:您可以通过让sumOfArray接受数组和起始(或结束)索引来简化事情。|1622475|entityMap^0|3|6|C|3|H|A|V|3|0|9|A|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|I|$]]

The issue is that <code>a[n-1]</code> is an <code>int</code>, whereas <code>sumOfArray</code> expects an array of <code>int</code>.

Hint: you can simplify things by making <code>sumOfArray</code> take the array and the starting (or ending) index.

blocks|key|1591142|text|a是一个int数组。因此，a[n-1]是一个int。您正在将一个int传递给sumOfArray，它需要一个数组，而不是一个int。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1591143|entityMap^0|0|1|4|3|D|6|M|3|W|3|12|A|1Q|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|G|$]]

<code>a</code> is an <code>int</code> array. Thus <code>a[n-1]</code> is an <code>int</code>. You are passing an <code>int</code> to <code>sumOfArray</code> which expects an array and not an <code>int</code>.

blocks|key|1622491|text|a[n-1]+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1622492|是在n-1处得到int，而不是从0到n-1的数组。|unstyled|1622493|试着使用|1622494|Arrays.copyOf(a,+a.length-1);|1622495|相反，|1622496|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$B|C]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>a[n-1] 
</code></pre>

is getting the int at n-1, not the array from 0 to n-1.

try using 

<pre><code>Arrays.copyOf(a, a.length-1);
</code></pre>

instead

blocks|key|810737|text|解决方案比它看上去更简单，请尝试如下(假设数组的长度为非零)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|810738|public+int+sumOfArray(int[]+a,+int+n)+{
++++if+(n+==+0)
++++++++return+a[n];
++++else
++++++++return+a[n]+%2B+sumOfArray(a,+n-1);
}|code-block|syntax|javascript|810739|就这样说吧：|810740|int[]+a+=+{+1,+2,+3,+4,+5+};
int+sum+=+sumOfArray(a,+a.length-1);|810741|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

The solution is simpler than it looks, try this (assuming an array with non-zero length):

<pre><code>public int sumOfArray(int[] a, int n) {
 if (n == 0)
 return a[n];
 else
 return a[n] + sumOfArray(a, n-1);
}
</code></pre>

Call it like this:

<pre><code>int[] a = { 1, 2, 3, 4, 5 };
int sum = sumOfArray(a, a.length-1);
</code></pre>

blocks|key|810754|text|如果不想传递数组的长度，请尝试这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|810755|private+static+int+sumOfArray(int[]+array)+{

++++++++if+(1+==+array.length)+{
++++++++++++return+array[array.length+-+1];
++++++++}

++++++++return+array[0]+%2B+sumOfArray(Arrays.copyOfRange(array,+1,+array.length));
++++}|code-block|syntax|javascript|810756|当然，您需要检查数组是否为空。|810757|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Try this if you don't want to pass the length of the array :

<pre><code>private static int sumOfArray(int[] array) {

 if (1 == array.length) {
 return array[array.length - 1];
 }

 return array[0] + sumOfArray(Arrays.copyOfRange(array, 1, array.length));
 }
</code></pre>

Offcourse you need to check if the array is empty or not.

blocks|key|810766|text|--这是一个复杂度为O(N).and的递归解决方案，只包含输入参数A[]+.|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|810767|在此解决方案中，可以将空和空(0长度)大小写具体处理为返回0。在这种情况下，您也会抛出异常。|810768|810769|/*
+*+Complexity+is+O(N)
+*/
public+int+recursiveSum2(int+A[])
{
++++if(A+==+null+%7C%7C+A.length==0)
++++{
++++++++return+0;
++++}
++++else
++++{
++++++++return+recursiveSum2Internal(A,A.length-1);
++++}
}
private+int+recursiveSum2Internal(int+A[],int+length)
{
++++if(length+==0+)
++++{
++++++++return+A[length];
++++}
++++else
++++{
++++++++return+A[length]%2BrecursiveSum2Internal(A,+length-1);
++++}
}|code-block|syntax|javascript|810770|entityMap^0|0|12|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|T|8|@]|D|@]|E|$]]|$1|I|3|J|5|K|7|U|8|@]|D|@]|E|$L|M]]|$1|N|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|O|$]]

This is the one recursive solution with complexity O(N).and with input parameter A[] only. 
You can handle null and empty(0 length) case specifically as Its returning 0 in this solution. You throw Exception as well in this case.

<hr>

<pre><code>/*
 * Complexity is O(N)
 */
public int recursiveSum2(int A[])
{
 if(A == null || A.length==0)
 {
 return 0;
 }
 else
 {
 return recursiveSum2Internal(A,A.length-1);
 }
}
private int recursiveSum2Internal(int A[],int length)
{
 if(length ==0 )
 {
 return A[length];
 }
 else
 {
 return A[length]+recursiveSum2Internal(A, length-1);
 }
}
</code></pre>

blocks|key|1591336|text|这个递归的解决方案怎么样？您可以创建一个较小的子数组，其中包含从第二个到末尾的元素。这种递归一直持续到数组大小变为1为止。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1591337|import+java.util.Arrays;

public+class+Sum+{
++++public+static+void+main(String[]+args){
++++++++int[]+arr+=+{1,2,3,4,5};
++++++++System.out.println(sum(arr));+//+15
++++}

++++public+static+int+sum(int[]+array){
++++++++if(array.length+==+1){
++++++++++++return+array[0];
++++++++}

++++++++int[]+subArr+=+Arrays.copyOfRange(array,+1,+array.length);
++++++++return+array[0]+%2B+sum(subArr);
++++}
}|code-block|syntax|javascript|1591338|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

How about this recursive solution? You make a smaller sub-array which contains elements from the second to the end. This recursion continues until the array size becomes 1.

<pre><code>import java.util.Arrays;

public class Sum {
 public static void main(String[] args){
 int[] arr = {1,2,3,4,5};
 System.out.println(sum(arr)); // 15
 }

 public static int sum(int[] array){
 if(array.length == 1){
 return array[0];
 }

 int[] subArr = Arrays.copyOfRange(array, 1, array.length);
 return array[0] + sum(subArr);
 }
}
</code></pre>

blocks|key|1622688|text|private+static+int+sum(int[]+arr)+{
++++//+TODO+Auto-generated+method+stub
++++int+n+=+arr.length;

++++if(n==1)
++++{
++++++++return+arr[n-1];
++++}

++++int+ans+=+arr[0]%2Bsum(Arrays.copyOf(arr,+n-1));

++++return+ans;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1622689|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>private static int sum(int[] arr) {
 // TODO Auto-generated method stub
 int n = arr.length;

 if(n==1)
 {
 return arr[n-1];
 }

 int ans = arr[0]+sum(Arrays.copyOf(arr, n-1));

 return ans;
}
</code></pre>

blocks|key|1622694|text|简化版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1622695|//acc+->+result+accumlator,+len+-+current+length+of+array

public+static+int+sum(int[]+arr,+int+len,+int+acc)+{
++++return+len+==+0+?+acc+:++sum(arr,+len-1,++arr[len-1]%2B+acc);+
}+++
public+static+void+main(String[]+args)++{
++++int[]+arr=+{+5,+1,+6,+2};
++++System.out.println(sum(arr,+arr.length,+0));
}|code-block|syntax|javascript|1622696|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Simplified version:

<pre><code>//acc -&gt; result accumlator, len - current length of array

public static int sum(int[] arr, int len, int acc) {
 return len == 0 ? acc : sum(arr, len-1, arr[len-1]+ acc); 
} 
public static void main(String[] args) {
 int[] arr= { 5, 1, 6, 2};
 System.out.println(sum(arr, arr.length, 0));
}
</code></pre>

blocks|key|1622703|text|没有任何预定义的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1622704|公共静态整数和(+int+input[])+{|1622705|+int+n+=+input.length;

++if+(n+==+0)++//+base+case
++return+0;
++
++++int+small[]=new+int[n-1];
++++
++++for(int+i=1;i<n;i%2B%2B)
++++{
++++++++small[i-1]=input[i];
++++}

++++++return+input[0]%2Bsum(small);
++++
}|code-block|syntax|javascript|1622706|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

without any predefined function.
public static int sum(int input[]) {
<pre><code> int n = input.length;

 if (n == 0) // base case
 return 0;
 
 int small[]=new int[n-1];
 
 for(int i=1;i&lt;n;i++)
 {
 small[i-1]=input[i];
 }

 return input[0]+sum(small);
 
}
</code></pre>

I have a program that I'm trying to make for class that returns the sum of all the integers in an array using recursion. Here is my program thus far:

<pre><code>public class SumOfArray {

private int[] a;
private int n;
private int result;

 public int sumOfArray(int[] a) {

 this.a = a;
 n = a.length;

 if (n == 0) // base case
 result = 0;
 else
 result = a[n] + sumOfArray(a[n-1]);

 return result;

 } // End SumOfArray method

} // End SumOfArray Class 
</code></pre>

But I'm getting three error which are all related, I believe, but I can't figure out why it is finding a type of null:

<pre><code>SumOfArray.java:25: sumOfArray(int[]) in SumOfArray cannot be applied to (int)
 result = a[n] + sumOfArray(a[n-1]);
 ^
SumOfArray.java:25: operator + cannot be applied to int,sumOfArray
 result = a[n] + sumOfArray(a[n-1]);
 ^
SumOfArray.java:25: incompatible types
found : &lt;nulltype&gt;
required: int
 result = a[n] + sumOfArray(a[n-1]);
 ^
3 errors
</code></pre>

recursively sum the integers in an array

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个程序，我试图为类创建一个程序，它使用递归返回数组中所有整数的和。到目前为止，我的计划如下：public class SumOfArray {private int[] a;private int n;private int result;    public int sumOfArray(int[] a) { ...

问递归地将数组中的整数求和
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问递归地将数组中的整数求和EN