blocks|key|2908013|text|编辑:我把东西“颠倒了”--现在修复了！|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2908014|是。按索引递增顺序遍历n个数组项，始终只将和添加到应添加到的下一个最小索引，而不是所有项：|2908015|for+i+=+1+to+n:
++++j+=+i+%2B+(i+&+-i)+++++#+Finds+next+higher+index+that+this+value+should+contribute+to
++++if+j+<=+n:
++++++++x[j]+%2B=+x[i]|code-block|syntax|javascript|2908016|这之所以有效，是因为尽管每个值都会贡献几个范围和，但在处理了该值所贡献的最低范围和之后(实际上不需要“处理”，因为和已经在那里了)，我们不再需要维护其单独的身份--它可以安全地与构成剩余范围和的所有其他值合并。|2908017|TTBOMK这个算法是“新的”--但是我还没仔细研究过;)|2908018|entityMap^0|0|K|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|Y|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

[EDIT: I had things "upside-down" -- fixed now!]

Yes. Loop through the n array items in increasing index order, always adding the sum only to the next smallest index that it should be added to, instead of to all of them:

<pre><code>for i = 1 to n:
 j = i + (i &amp; -i) # Finds next higher index that this value should contribute to
 if j &lt;= n:
 x[j] += x[i]
</code></pre>

This works because although every value contributes to several range sums, after processing the bottommost range sum that the value contributes to (which actually requires no "processing", since the sum is already in there), we no longer need to maintain its separate identity -- it can safely be merged with all other values that contribute to the remaining range sums.

TTBOMK this algorithm is "new" -- but then I haven't looked very hard ;)

blocks|key|2695836|text|下面是Java实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2695837|public+BIT(long[]+nums)+{
++++++++bit+=+new+long[nums.length+%2B+1];+//one-indexed
++++++++for+(int+i+=+1;+i+<=+nums.length;+i%2B%2B)+{
++++++++++++bit[i]+%2B=+nums[i+-+1];+//update+node
++++++++++++if+(i+%2B+(i+&+-i)+<=+nums.length)+{
++++++++++++++++bit[i+%2B+(i+&+-i)]+%2B=+bit[i];+//update+parent
++++++++++++}
++++++++}
++++}|code-block|syntax|javascript|2695838|与j_random_hacker的帖子的一般思想相同:我们更新当前节点和下一个更高的父节点，使用所有子节点总是在其各自的父节点之前被访问的属性|2695839|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|$]]

Here's the Java implementation:
<pre><code>public BIT(long[] nums) {
 bit = new long[nums.length + 1]; //one-indexed
 for (int i = 1; i &lt;= nums.length; i++) {
 bit[i] += nums[i - 1]; //update node
 if (i + (i &amp; -i) &lt;= nums.length) {
 bit[i + (i &amp; -i)] += bit[i]; //update parent
 }
 }
 }
</code></pre>
Same general idea as j_random_hacker's post: we update the current node and the next higher parent node, using the property that all children nodes will always be visited before their respective parent nodes

<a href="https://en.wikipedia.org/wiki/Fenwick_tree" rel="noreferrer">Fenwick tree</a> is a data structure which allows two kind of operations (you can augment it with more operations):

<ul>
<li>point update <code>update(index, value)</code></li>
<li>prefix sum <code>query(index)</code></li>
</ul>

Both of the operations are in <code>O(log(n))</code> where <code>n</code> is the size of an array. I have no problems understanding how to do both operations and the logic behind them.

<hr>

My question is how can I initialize a Fenwick tree from an array. Clearly I can achieve this in <code>O(nlog(n))</code>, by calling <code>n</code> times <code>update(i, arr[i])</code>, but is there a way to initialize it in <code>O(n)</code>.

<hr>

Why am I asking this if wikipedia tells that you can initialize in <code>nlog(n)</code>? Because the article is so rudimentary, that I am not sure whether it is the best complexity one can achieve. Also drawing parallels with naive heap creation which is done by populating the heap one by one and can be achieved in <code>O(nlog(n))</code> versus smart heap initialization in <code>O(n)</code> gives me hope that something similar can be done in Fenwick tree.

Is it possible to build a Fenwick tree in O(n)?

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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是一种数据结构，它允许两种操作(你可以用更多的操作来扩充它)：update(index, value)prefix sum query(index)的点更新这两个操作都是在O(log(n))中进行的，其中n是数组的大小。我完全理解如何进行这两种操作以及它们背后的逻辑。我的问题是如何从数组初始化Fenwick树。显然，我...

问在O(n)中构建Fenwick树是可能的吗？
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