blocks|key|4193875|text|至少有两种方法可以解决你的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4193876|第一个示例使用循环来计算峰值总和：|4193877|int+peak_sum(int+arr[],+int+size)
{
++++int+sum+=+0;

++++//+check+first+element
++++if+(arr[0]+>+arr[1])+{
++++++++std::cout+<<+arr[0]+<<+std::endl;
++++++++sum+%2B=+arr[0];
++++}

++++//+for+each+middle+value+in+array
++++for+(int+i+=+1;+i+<+size+-+2;+i%2B%2B)+{
++++++++//+if+current+value+is+peak
++++++++if(arr[i]+>+arr[i+%2B+1]+&&+arr[i]+>+arr[i+-+1])+{
++++++++++++std::cout+<<+arr[i]+<<+std::endl;
++++++++++++//+then+we+add+it+to+total+sum+of+peaks
++++++++++++sum+%2B=+arr[i];
++++++++}
++++}

++++//+check+last+element
++++if+(arr[size+-+1]+>+arr[size+-+2])+{
++++++++std::cout+<<+arr[size+-+1]+<<+std::endl;
++++++++sum+%2B=+arr[size+-+1];
++++}++++

++++//+after+all+we+return+this+sum+of+peaks
++++return+sum;
}|code-block|syntax|javascript|4193878|但是如果你想以递归的方式解决它，那么它可能看起来像这样：|4193879|int
peak_sum(int+arr[],+int+size,+int+index,+int+sum)
{
++++//+if+index+in+bounds
++++if+(index+<+size)+{
++++++++//+if+we+on+the+first+element
++++++++if+(index+==+0)+{
++++++++++++//+if+first+element+more+then+next:+it+is+peak
++++++++++++if+(arr[index]+>+arr[index+%2B+1])+{
++++++++++++++++std::cout+<<+arr[index]+<<+std::endl;
++++++++++++++++sum+%2B=+arr[index];
++++++++++++}

++++++++++++//+if+we+on+the+last+element
++++++++}+else+if+(index+==+size+-+1)+{
++++++++++++//+if+first+element+more+then+previous:+it+is+peak
++++++++++++if+(arr[index]+>+arr[index+-+1])+{
++++++++++++++++std::cout+<<+arr[index]+<<+std::endl;
++++++++++++++++sum+%2B=+arr[index];
++++++++++++}

++++++++++++//+else+we+in+the+middle
++++++++}+else+{
++++++++++++//+if+current+element+more+then+previous+and+next:+it+is+peak
++++++++++++if+(arr[index]+>+arr[index+-+1]+&&+arr[index]+>+arr[index+%2B+1])+{
++++++++++++++++std::cout+<<+arr[index]+<<+std::endl;
++++++++++++++++sum+%2B=+arr[index];
++++++++++++}
++++++++}

++++++++//+anyway+we+check+next+value+of+array
++++++++//+by+incrimeting+current+index
++++++++return+peak_sum(arr,+size,+index+%2B+1,+sum);
++++}+else+{
++++++++//+otherwise+we+end+iterating+array
++++++++return+sum;
++++}
}|4193880|并使用以下命令调用它：|4193881|//+++++++array,+size+of+it,+initial+index,+initial+sum
peak_sum(arr,+array_size,+0,+0);|4193882|如果你对你的代码流感兴趣：|4193883|//example+for+array_size+==+3

if+(index+in+bounds)+{
++++index%2B%2B;
++++if+(index+in+bounds)+{
++++++++index%2B%2B;
++++++++if+(index+in+bounds)+{
++++++++++++index%2B%2B;
++++++++++++if+(index+in+bounds)+{}
++++++++++++else+return+sum;
++++++++}
++++}
}|4193884|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

There are at least 2 ways to solve your problem.

First example uses loops for calculating sum of peaks:

<pre><code>int peak_sum(int arr[], int size)
{
 int sum = 0;

 // check first element
 if (arr[0] &gt; arr[1]) {
 std::cout &lt;&lt; arr[0] &lt;&lt; std::endl;
 sum += arr[0];
 }

 // for each middle value in array
 for (int i = 1; i &lt; size - 2; i++) {
 // if current value is peak
 if(arr[i] &gt; arr[i + 1] &amp;&amp; arr[i] &gt; arr[i - 1]) {
 std::cout &lt;&lt; arr[i] &lt;&lt; std::endl;
 // then we add it to total sum of peaks
 sum += arr[i];
 }
 }

 // check last element
 if (arr[size - 1] &gt; arr[size - 2]) {
 std::cout &lt;&lt; arr[size - 1] &lt;&lt; std::endl;
 sum += arr[size - 1];
 } 

 // after all we return this sum of peaks
 return sum;
}
</code></pre>

But if you want to solve it in a recursive way then it may looks like this:

<pre><code>int
peak_sum(int arr[], int size, int index, int sum)
{
 // if index in bounds
 if (index &lt; size) {
 // if we on the first element
 if (index == 0) {
 // if first element more then next: it is peak
 if (arr[index] &gt; arr[index + 1]) {
 std::cout &lt;&lt; arr[index] &lt;&lt; std::endl;
 sum += arr[index];
 }

 // if we on the last element
 } else if (index == size - 1) {
 // if first element more then previous: it is peak
 if (arr[index] &gt; arr[index - 1]) {
 std::cout &lt;&lt; arr[index] &lt;&lt; std::endl;
 sum += arr[index];
 }

 // else we in the middle
 } else {
 // if current element more then previous and next: it is peak
 if (arr[index] &gt; arr[index - 1] &amp;&amp; arr[index] &gt; arr[index + 1]) {
 std::cout &lt;&lt; arr[index] &lt;&lt; std::endl;
 sum += arr[index];
 }
 }

 // anyway we check next value of array
 // by incrimeting current index
 return peak_sum(arr, size, index + 1, sum);
 } else {
 // otherwise we end iterating array
 return sum;
 }
}
</code></pre>

And call it with:

<pre><code>// array, size of it, initial index, initial sum
peak_sum(arr, array_size, 0, 0);
</code></pre>

If you interested in flow of your code:

<pre><code>//example for array_size == 3

if (index in bounds) {
 index++;
 if (index in bounds) {
 index++;
 if (index in bounds) {
 index++;
 if (index in bounds) {}
 else return sum;
 }
 }
}
</code></pre>

I am not able to understand the flow of program in recursion with and without return both.How do i return or print the group of values calculated by the function?For eg To calculate all the peak elements in an array I used recursion but I am not able to get how to give those values as a result.Basically I am not clear what happens if i write or not write return before the recursive function.

<pre><code>int peak(int arr[],int i,int size)
{
 while(i&lt;size) 
 {
 if(arr[i]&gt;arr[i+1]&amp;&amp;arr[i]&gt;arr[i-1])
 cout&lt;&lt;arr[i];
 i++;
 return peak(arr,i,size);
 }
}
</code></pre>

Return in Recursion and basic flow

我不能理解有或没有return的递归程序的流程，我是返回还是打印函数计算出的一组值?例如，为了计算一个数组中的所有峰值元素，我使用了递归，但我不能得到如何将这些值作为result.Basically给出，我不清楚如果我在递归函数之前编写或不编写return会发生什么。int  peak(int arr[],int i,...

问递归和基本流程中的返回
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