blocks|key|5993825|text|为什么要使用两个循环？如果数字为13，则不要将其或下一个数字添加到总和中，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5993826|public+int+sum13(int[]+nums)+{
++++int+sum+=+0;
++++for+(int+i+=+0;+i+<+nums.length;+i%2B%2B)+{
++++++++if+(nums[i]+==+13)+{
++++++++++++i%2B%2B;
++++++++}+else+{
++++++++++++sum+=+sum+%2B+nums[i];
++++++++}
++++}
++++return+sum;
}|code-block|syntax|javascript|5993827|entityMap^0|G|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Why are you using two loops? If the number is <code>13</code>, just don't add it or the next number to the sum, like this: 

<pre><code>public int sum13(int[] nums) {
 int sum = 0;
 for (int i = 0; i &lt; nums.length; i++) {
 if (nums[i] == 13) {
 i++;
 } else {
 sum = sum + nums[i];
 }
 }
 return sum;
}
</code></pre>

blocks|key|2641644|text|根据你的问题，如果数字是13+&它的直接后继者，则不包括在内。因此，当您发现13再次递增循环计数器时：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2641645|+for(int+i=0;i<nums.length;i%2B%2B){
+++if(nums[i]+==+13)+{
+++++i%2B%2B;
++++}
+++else+{
+++++sum+=+sum+%2B+nums[i];
+++}
++}|code-block|syntax|javascript|2641646|如果数组中有一个元素，这自然会跳过13的下一个元素。|2641647|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|$]]

As per your question you exclude if the number is 13 &amp; its immediate successor. So when you find 13 increment the loop counter once more:

<pre><code> for(int i=0;i&lt;nums.length;i++){
 if(nums[i] == 13) {
 i++;
 }
 else {
 sum = sum + nums[i];
 }
 }
</code></pre>

This will naturally skip the next element of 13, if there's one within the array.

blocks|key|1085737|text|直接在第一个for循环中求和如何？更短、更容易、更快的方式:)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1085738|public+int+sum13(int[]+nums)+{
++int+sum+=+0;

++for+(int+i=0;i<nums.length;i%2B%2B){
++++if+(nums[i]+!=+13)+{
++++++sum+%2B=+nums[i];
++++}+else+{
++++++i%2B%2B;
++++}
++}
++return+sum;
}|code-block|syntax|javascript|1085739|entityMap^0|6|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

How about summing directly in the first <code>for</code> loop? Shorter, easier and faster way :)

<pre><code>public int sum13(int[] nums) {
 int sum = 0;

 for (int i=0;i&lt;nums.length;i++){
 if (nums[i] != 13) {
 sum += nums[i];
 } else {
 i++;
 }
 }
 return sum;
}
</code></pre>

blocks|key|1085700|text|你听说continue做什么了吗？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1085701|if(nums[i]+==+13){
+++++d+=+i;
+++++continue;
}|code-block|syntax|javascript|1085702|entityMap^0|3|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Did you hear what <code>continue</code> does?

<pre><code>if(nums[i] == 13){
 d = i;
 continue;
}
</code></pre>

blocks|key|517309|text|public+int+sum13(int[]+nums)+{
++++++boolean+skip+=+false;
++++++int+sum+=+0;

++++++for+(int+i=0;i<nums.length;i%2B%2B){
++++++++if(nums[i]+==+13){
++++++++++++skip=true;
++++++++++++continue;
++++++++}
++++++++else{
++++++++if(skip)+{
++++++++++++skip=false;
++++++++++++continue;
++++++++}
++++++++sum%2B=nums[i];
++++++++}
++++++}

++++++return+sum;
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|517310|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>public int sum13(int[] nums) {
 boolean skip = false;
 int sum = 0;

 for (int i=0;i&lt;nums.length;i++){
 if(nums[i] == 13){
 skip=true;
 continue;
 }
 else{
 if(skip) {
 skip=false;
 continue;
 }
 sum+=nums[i];
 }
 }

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

blocks|key|517341|text|您实际上正在做的是在13的第一次出现时停止。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|517342|试试这个：|517343|public+int+sum13(int[]+nums)+{
++++int+sum+=+0;

++++for+(int+i+=+0;+i+<+nums.length;+i%2B%2B)+{
++++++++if+(nums[i]+==+13)+{
++++++++++++i%2B%2B;
++++++++}+else+{
++++++++++++sum+%2B=+nums[i];
++++++++}
++++}

++++return+sum;
}|code-block|syntax|javascript|517344|我没有测试它，但它应该可以工作...|517345|干杯!|517346|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

What you are actually doing is stopping at the first occurrence of 13.

try this:

<pre><code>public int sum13(int[] nums) {
 int sum = 0;

 for (int i = 0; i &lt; nums.length; i++) {
 if (nums[i] == 13) {
 i++;
 } else {
 sum += nums[i];
 }
 }

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

I didn't test it but it should work...

Cheers!

blocks|key|517348|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|517349|+++public+int+mySum(int[]+nums,+int+except)+{
++++++++int+sum+=+0;
++++++++for+(int+i+=+0;+i+<+nums.length;+i%2B%2B)+{
+++++++++++++sum+%2B=+(nums[i]+==+except)+?+0:+nums[i];
++++++++}
++++++++return+sum;
++++}|code-block|syntax|javascript|517350|然后|517351|mySum(nums,+13)|517352|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|$]]

Try this 

<pre><code> public int mySum(int[] nums, int except) {
 int sum = 0;
 for (int i = 0; i &lt; nums.length; i++) {
 sum += (nums[i] == except) ? 0: nums[i];
 }
 return sum;
 }
</code></pre>

Then 

<pre><code>mySum(nums, 13)
</code></pre>

blocks|key|517392|text|当值等于13时，使用continue；语句。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|517393|public+int+sum13(int[]+nums)+{
++int+sum+=+0;
++for+(int+i=0;i<nums.length;i%2B%2B){
++++if(nums[i]+==+13){
+++++++continue;
++++}
++++else{
+++++++sum+=+sum+%2B+nums[i];
++++}
++}
++return+sum;
}|code-block|syntax|javascript|517394|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|$]]

use the continue; statement when the value equals to 13.

<pre><code>public int sum13(int[] nums) {
 int sum = 0;
 for (int i=0;i&lt;nums.length;i++){
 if(nums[i] == 13){
 continue;
 }
 else{
 sum = sum + nums[i];
 }
 }
 return sum;
}
</code></pre>

blocks|key|1081820|text|尝尝这个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1081821|public+int+sum(int[]+nums)+{
++++int+sum+=+0;
++++for+(int+i+=+0;+i+<+nums.length;)+{
++++++++if+(nums[i]+==+13)
++++++++++++i+=+i+%2B+2;
++++++++else+{
++++++++++++sum+=+sum+%2B+nums[i];
++++++++++++i%2B%2B;
++++++++}
++++}
++++return+sum;
}|code-block|syntax|javascript|1081822|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|$]]

Try this.

<pre><code>public int sum(int[] nums) {
 int sum = 0;
 for (int i = 0; i &lt; nums.length;) {
 if (nums[i] == 13)
 i = i + 2;
 else {
 sum = sum + nums[i];
 i++;
 }
 }
 return sum;
}
</code></pre>

So, basically, the questions asks me to find the sum of the numbers in an array. Except the number '13' is very unlucky, so it does not count '13' and the number that come immediately after a '13'.

This is what I did:

<pre><code>public int sum13(int[] nums) {
 int d = 0;
 int sum = 0;

 for (int i=0;i&lt;nums.length;i++){
 if(nums[i] == 13){
 d = i;
 break;
 }
 else{
 d = nums.length;
 }
 }
 for(int i=0;i&lt;d;i++){
 sum = sum + nums[i];
 }
 return sum;
}
</code></pre>

Even though I pass most of the tests, I still can't understand how to exclude the number right next to the 13 from the sum.

For example, sum13({1, 2, 2, 1, 13}) → 6 PASSES!
sum13({13, 1, 13}) → 0 PASSES! 
But, sum13({13, 1, 2, 13, 2, 1, 13}) → 3 RETURNS 0 instead as it stops at the first instance of 13.

Sum of 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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所以，基本上，这些问题要求我找出数组中数字的总和。除了数字'13‘是非常不吉利的，所以它不计算'13’和紧跟在'13‘后面的数字。这是我所做的：public int sum13(int[] nums) { int d = 0; int sum = 0; for (int i=0;i<nums.length;i++){ if(nums[i] == 13){ d = i; br

问数组中整数的总和
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回答 9

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问数组中整数的总和EN

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问数组中整数的总和
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