blocks|key|71445|text|如果数组已排序，则可以直接比较连续元素的重复项，直到找到不同的长度为止。循环完成后，可以使用数组长度属性返回新的长度。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|71446|显然，你必须在某些地方管理你的逻辑才能做到这一点。|71447|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

if the array is sorted, you can directly compare consecutive elements for duplicates until you find a different element.Once the loop is completed you can return the new length using array length property.

Obviously you have to manage your logic at some places for doing the same.

blocks|key|1629098|text|我试着用下面的逻辑来解决这个问题。但是它不是数组分配的地方。仍然不确定如何从具有就地逻辑的数组中删除null。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1629099|+public+static+void+main(String[]+args)+{

++++++++++++Integer[]+initArray+=+{+2,+3,+4,+4,+5,+6,+7,+8,+8,+9,+10,+11+};
++++++++++++System.out.println("Initial+Array+:+"%2B+Arrays.toString(initArray));

++++++++++++System.out.println("Array+Size+After++Duplicates+Removal+:+"%2B+removeDuplicates(initArray));
++++++++}

++++public+static+int+removeDuplicates(Integer[]+arr)+{

++++++++Integer+prev+=+null;
++++++++for(int+index=+0;+index+<=+arr.length+-1;++index+%2B%2B){
++++++++++++if(arr[index]+==+prev)
++++++++++++{
++++++++++++++++remove(index,+arr);
++++++++++++++++prev+=+arr[index];
++++++++++++}else{
++++++++++++++++prev+=+arr[index];
++++++++++++++++continue;
++++++++++++}
++++++++}

++++++++List<Integer>+items+=+new+ArrayList<Integer>(arr.length);
++++++++for(Integer+input+:+arr)+{
+++++++++++if+(input+!=+null)+{
++++++++++++++items.add(input);
+++++++++++}
++++++++}
++++++++Integer[]+outputs+=+items.toArray(new+Integer[items.size()]);
++++++++System.out.println("Final+Array+:+"%2B+Arrays.toString(outputs));
++++++++return+outputs.length;
++++}

++++public+static+void+remove(int+index,+Integer[]+a){

++++++++int+lastIndex+=+a.length-1;
++++++++if(index+<=+lastIndex){
++++++++++++a[index]+=null;
++++++++++++int+tmp+=+index;
++++++++++++while(tmp+<+lastIndex){
++++++++++++++++a[tmp]+=+a[tmp%2B1];
++++++++++++++++a[tmp%2B1]+=+null;
++++++++++++++++tmp%2B%2B;
++++++++++++}
++++++++++++lastIndex--;
++++++++}+else+{
++++++++++++throw+new+ArrayIndexOutOfBoundsException();
++++++++}

++++}+|code-block|syntax|javascript|1629100|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|$]]

I tried to solve this with below logic. It is not in place array allocation though. Still not sure how to remove null from an array with in place logic.

<pre><code> public static void main(String[] args) {

 Integer[] initArray = { 2, 3, 4, 4, 5, 6, 7, 8, 8, 9, 10, 11 };
 System.out.println("Initial Array : "+ Arrays.toString(initArray));

 System.out.println("Array Size After Duplicates Removal : "+ removeDuplicates(initArray));
 }

 public static int removeDuplicates(Integer[] arr) {

 Integer prev = null;
 for(int index= 0; index &lt;= arr.length -1; index ++){
 if(arr[index] == prev)
 {
 remove(index, arr);
 prev = arr[index];
 }else{
 prev = arr[index];
 continue;
 }
 }

 List&lt;Integer&gt; items = new ArrayList&lt;Integer&gt;(arr.length);
 for(Integer input : arr) {
 if (input != null) {
 items.add(input);
 }
 }
 Integer[] outputs = items.toArray(new Integer[items.size()]);
 System.out.println("Final Array : "+ Arrays.toString(outputs));
 return outputs.length;
 }

 public static void remove(int index, Integer[] a){

 int lastIndex = a.length-1;
 if(index &lt;= lastIndex){
 a[index] =null;
 int tmp = index;
 while(tmp &lt; lastIndex){
 a[tmp] = a[tmp+1];
 a[tmp+1] = null;
 tmp++;
 }
 lastIndex--;
 } else {
 throw new ArrayIndexOutOfBoundsException();
 }

 } 
</code></pre>

I just started coding LeetCode problmes. The problem is the following:
Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

My solution is the following:

<pre><code>public static int removeDuplicates1(int[] A) {
 Set&lt;Integer&gt; mySet = new HashSet&lt;Integer&gt;();

 for(Integer i: A){
 mySet.add(i);
 }
 return mySet.size();
 }
</code></pre>

Even though I get the right answer in Eclipse, my answer was not accepted by the website. Can anyone tell me what is wrong?

Removing doubles from 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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我刚刚开始编写LeetCode problmes。问题如下:给定一个排序的数组，去掉重复的元素，使每个元素只出现一次，并返回新的长度。不要为另一个数组分配额外的空间，您必须使用常量内存在适当的位置执行此操作。我的解决方案如下：public static int removeDuplicates1(int[] A) { ...

问从数组中删除双精度
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