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    <p><strike>If you're using the Sun implementation, it's <code>O(log(n))</code>.</strike> From the <a href="http://java.sun.com/javase/6/docs/api/java/util/PriorityQueue.html" rel="noreferrer">Javadocs</a>:</p> <blockquote> <p>Implementation note: this implementation provides O(log(n)) time for the enqueing and dequeing methods (<code>offer</code>, <code>poll</code>, <code>remove()</code> and <code>add</code>); linear time for the <code>remove(Object)</code> and <code>contains(Object)</code> methods; and constant time for the retrieval methods (<code>peek</code>, <code>element</code>, and <code>size</code>).</p> </blockquote> <p>Other implementations could have different complexity.</p> <hr> <p><strong>Edit:</strong> The Javadocs don't cover the performance of removing an element with an iterator, so I had to look up the source code. This is all relevant to the Sun implementation, and may differ in Apple's version, GNU Classpath, etc. Sun's source is available <a href="http://download.java.net/jdk6/" rel="noreferrer">here</a>; it is also included in the JDK, so you might already have it installed.</p> <p>In the <code>PriorityQueue</code>'s iterator, the default case for <code>remove()</code> is to call <code>PriorityQueue.removeAt(lastRet)</code>, where <code>lastRet</code> is the index that was last returned by <code>next()</code>. <code>removeAt()</code> appears to be <code>O(log(n))</code> worst case (it might have to sift the queue, but doesn't have to iterate).</p> <p>However, sometimes bad things happen. From the comments of <code>removeAt()</code>:</p> <pre><code>/** * Removes the ith element from queue. * * Normally this method leaves the elements at up to i-1, * inclusive, untouched. Under these circumstances, it returns * null. Occasionally, in order to maintain the heap invariant, * it must swap a later element of the list with one earlier than * i. Under these circumstances, this method returns the element * that was previously at the end of the list and is now at some * position before i. This fact is used by iterator.remove so as to * avoid missing traversing elements. */ </code></pre> <p>When a non-null element is returned by <code>removeAt()</code>, the iterator adds it to a special queue for later use: when the iterator runs out of elements in the queue, it then iterates through this special queue. When <code>remove()</code> is called during this second phase of iteration, the iterator calls <code>PriorityQueue.removeEq(lastRetElt)</code>, where <code>lastRetElt</code> is the last element returned from the special queue. <code>removeEq</code> is forced to use a linear search to find the correct element to remove, which makes it <code>O(n)</code>. BUT it can check elements using <code>==</code> rather than <code>.equals()</code>, so its constant factor is lower than <code>PriorityQueue.remove(Object)</code>.</p> <p>So, in other words, removing with an iterator is technically <code>O(n)</code>, but in practice it should be quite a bit faster than <code>remove(Object)</code>.</p>
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    1. POJava PriorityQueue removal of arbitrary elements performance
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    1. COIt doesn't specify the remove() operation time from an iterator. Removing the head will definitely take O(log(n)), but what about removing other elements?
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    2. COGood question. Let me look at the source; it might use `remove(Object)`, which would make it linear.
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    3. COremove() uses indexOf(), so it is linear
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