1 首先要看你的List是怎么生成的，比如：List<String> strList = Arrays.asList("a", "b", "aa", "ab", "ba");这种方式生成的List是不能改变的（fixed size），具体可以参见源码。2 比如下面这种方式生成的List是可以改变的：List<String> strList2 = new ArrayList<>();strList2.add("a");strLi...
The following Java program usesList.removeIf()to remove multiple elements from the arraylistin java by element value. ArrayList<String>namesList=newArrayList<String>(Arrays.asList("alex","brian","charles","alex"));System.out.println(namesList);namesList.removeIf(name->name.equals("alex"));Syst...
因为要求是返回修改后的长度并只考虑该长度的数组,那么就不用考虑该长度之后的数组,所以只需得到索引 j 的值,不用再把索引 j 的值改为索引 i的值。 Java: 代码语言:txt AI代码解释 class Solution { public int removeElement(int[] nums, int val) { int i=0,j=nums.length-1;//i-左指针;j-右指...
} }returnlist.size() ; //返回数组length } 方法二:采用两个指针,不需要额外空间,数组原地做修改 publicintremoveElement(int[] nums,intval) {//原地修改,不需要额外的空间intnewindex = 0;for(inti = 0; i < nums.length; i++) {if(nums[i] !=val) nums[newindex++] =nums[i]; }returnnewinde...
checkElementIndex(index); return node(index).item; } Node node(int index) { if (index < (size >> 1)) { Node x = first; for (int i = 0; i < index; i++) x = x.next; return x; } else { Node x = last; for (int i = size - 1; i > index; i--) ...
因此新的方法更适合容易出现异常条件的情况。 peek,element区别: element() 和 peek() 用于在队列的头部查询元素。与 remove() 方法类似,在队列为空时, element() 抛出一个异常,而 peek() 返回 null。
* Removes the element at the specified position in this list. * Shifts any subsequent elements to the left (subtracts one from their * indices). * * @param index the index of the element to be removed * @return the element that was removed from the list ...
leecode-数组-27Remove Element-java 27.Remove Element https://leetcode.com/problems/remove-element/ 看起来很简单的样子,甚至准备直接写。还是图样图森破,很多地方出了岔子。改了好半天。警醒自己!!总是想不全面。有点小难过。 1 2 3 4 5 6
if ( s.equals(in) ) { list.remove(s); } } } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 看上去我们的代码很完美,在理论上肯定会达到预期的结果,但是当遍历删除后的array,会发现,打印结果为2,2,1,3,3。怎么还有一个1没删除呢,现在我们来看在这段代码中用到的remove方法 ...
1. UsingCollection.removeIf()to Remove Duplicates from OriginalList TheremoveIf()method removes all of the elements of this collection that satisfy a specifiedPredicate. Each matching element is removed usingIterator.remove(). If the collection’s iterator does not support removal, then anUnsupportedOp...