Question

我正在阅读一些算法书并遇到过这一行，

Heaps are well suited for algorithms that merge sorted data streams.

没有任何解释说明为什么会这样。有人可以帮我解释一下为什么会这样吗？

Answer 1

如果您只有两个数据流，那么您真的不需要堆，如以下算法所示：

Let s1 ans s2 be the two streams

while s1.hasData() and s2.hasData()
    if s1.peek() < s2.peek(): datum = s1.pop()
    else: datum = s2.pop()
    s.push(datum)
if either is non-empty (only one is), add the rest of its content to s

正如Henk Holterman指出的那样，如果你有k>2个流，那么你就可以通过堆实现合并（基本上堆可以做出当前步骤中使用哪个流的现在复杂的决定）：

let H be a (min  or max, depending on your needs) heap
let s1, s2, ..., sk be sorted streams

// fill the heap with the first elements from the streams (e.g. min/max elements from each stream, depending on how they're sorted)
for i=1 to k:
    H.add((i, si.pop())) // we need to know which stream the element came from

let s be the initially-empty data stream which will contain the merged content in sorted order
// H.empty() will indicate that all streams are empty
while not H.empty():
    // take the min/max element of the min/max elements of each stream (*the* min/max element)
    (i, datum) = H.extract()
    // add it to s
    s.push(datum)
    // we know the datum came from s[i]; thus we need to push the next element from the i-th stream into heap as it may contain the next min/max element (that is, if s[i] isn't empty)
    if not s[i].empty():
        // we'll assume the heap sorts based on the second component of the pair
        H.add((i, s[i].pop())
// s is the sorted stream containing elements from s1,s2,...,sk

此运行时间为O((|s1|+...+|sk|) * log k)，其中|si|表示流si中元素的数量。

关键的想法是，在while - 循环的每次迭代中，您将s[1].peek()，s[2].peek()，...，s[k].peek()的最小/最大值添加到{ {1}}。您可以使用堆来实现此目的，该堆告诉您当前包含最小/最大元素的流。并注意如何优雅地处理流为空的边缘情况。

为什么堆非常适合合并排序的流？

1 个答案: