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Fix heap.md

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Yudong Jin
2023-01-31 19:11:48 +08:00
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@ -10,7 +10,7 @@ comments: true
- 「小顶堆 Min Heap」，任意结点的值 $\leq$ 其子结点的值； - 「小顶堆 Min Heap」，任意结点的值 $\leq$ 其子结点的值；
<figure markdown> <figure markdown>
![min_heap_and_max_heap](heap.assets/min_heap_and_max_heap.png){ width="600" } ![min_heap_and_max_heap](heap.assets/min_heap_and_max_heap.png)
</figure> </figure>
## 8.1.1. 堆术语与性质 ## 8.1.1. 堆术语与性质
@ -217,9 +217,7 @@ comments: true
具体地，给定索引 $i$ ，那么其左子结点索引为 $2i + 1$ 、右子结点索引为 $2i + 2$ 、父结点索引为 $(i - 1) / 2$ （向下整除）。当索引越界时，代表空结点或结点不存在。 具体地，给定索引 $i$ ，那么其左子结点索引为 $2i + 1$ 、右子结点索引为 $2i + 2$ 、父结点索引为 $(i - 1) / 2$ （向下整除）。当索引越界时，代表空结点或结点不存在。
<figure markdown> ![representation_of_heap](heap.assets/representation_of_heap.png)
![representation_of_heap](heap.assets/representation_of_heap.png){ width="600" }
</figure>
我们将索引映射公式封装成函数，以便后续使用。 我们将索引映射公式封装成函数，以便后续使用。
@ -419,23 +417,23 @@ comments: true
=== "Step 1" === "Step 1"
<figure markdown> <figure markdown>
![heap_push_step1](heap.assets/heap_push_step1.png){ width="600" } ![heap_push_step1](heap.assets/heap_push_step1.png)
</figure> </figure>
=== "Step 2" === "Step 2"
![heap_push_step2](heap.assets/heap_push_step2.png){ width="600" } ![heap_push_step2](heap.assets/heap_push_step2.png)
=== "Step 3" === "Step 3"
![heap_push_step3](heap.assets/heap_push_step3.png){ width="600" } ![heap_push_step3](heap.assets/heap_push_step3.png)
=== "Step 4" === "Step 4"
![heap_push_step4](heap.assets/heap_push_step4.png){ width="600" } ![heap_push_step4](heap.assets/heap_push_step4.png)
=== "Step 5" === "Step 5"
![heap_push_step5](heap.assets/heap_push_step5.png){ width="600" } ![heap_push_step5](heap.assets/heap_push_step5.png)
=== "Step 6" === "Step 6"
![heap_push_step6](heap.assets/heap_push_step6.png){ width="600" } ![heap_push_step6](heap.assets/heap_push_step6.png)
设结点总数为 $n$ ，则树的高度为 $O(\log n)$ ，易得堆化操作的循环轮数最多为 $O(\log n)$ ，**因而元素入堆操作的时间复杂度为 $O(\log n)$** 。 设结点总数为 $n$ ，则树的高度为 $O(\log n)$ ，易得堆化操作的循环轮数最多为 $O(\log n)$ ，**因而元素入堆操作的时间复杂度为 $O(\log n)$** 。
@ -570,34 +568,34 @@ comments: true
顾名思义，**从顶至底堆化的操作方向与从底至顶堆化相反**，我们比较根结点的值与其两个子结点的值，将最大的子结点与根结点执行交换，并循环以上操作，直到越过叶结点时结束，或当遇到无需交换的结点时提前结束。 顾名思义，**从顶至底堆化的操作方向与从底至顶堆化相反**，我们比较根结点的值与其两个子结点的值，将最大的子结点与根结点执行交换，并循环以上操作，直到越过叶结点时结束，或当遇到无需交换的结点时提前结束。
=== "Step 1" === "Step 1"
![heap_poll_step1](heap.assets/heap_poll_step1.png){ width="600" } ![heap_poll_step1](heap.assets/heap_poll_step1.png)
=== "Step 2" === "Step 2"
![heap_poll_step2](heap.assets/heap_poll_step2.png){ width="600" } ![heap_poll_step2](heap.assets/heap_poll_step2.png)
=== "Step 3" === "Step 3"
![heap_poll_step3](heap.assets/heap_poll_step3.png){ width="600" } ![heap_poll_step3](heap.assets/heap_poll_step3.png)
=== "Step 4" === "Step 4"
![heap_poll_step4](heap.assets/heap_poll_step4.png){ width="600" } ![heap_poll_step4](heap.assets/heap_poll_step4.png)
=== "Step 5" === "Step 5"
![heap_poll_step5](heap.assets/heap_poll_step5.png){ width="600" } ![heap_poll_step5](heap.assets/heap_poll_step5.png)
=== "Step 6" === "Step 6"
![heap_poll_step6](heap.assets/heap_poll_step6.png){ width="600" } ![heap_poll_step6](heap.assets/heap_poll_step6.png)
=== "Step 7" === "Step 7"
![heap_poll_step7](heap.assets/heap_poll_step7.png){ width="600" } ![heap_poll_step7](heap.assets/heap_poll_step7.png)
=== "Step 8" === "Step 8"
![heap_poll_step8](heap.assets/heap_poll_step8.png){ width="600" } ![heap_poll_step8](heap.assets/heap_poll_step8.png)
=== "Step 9" === "Step 9"
![heap_poll_step9](heap.assets/heap_poll_step9.png){ width="600" } ![heap_poll_step9](heap.assets/heap_poll_step9.png)
=== "Step 10" === "Step 10"
![heap_poll_step10](heap.assets/heap_poll_step10.png){ width="600" } ![heap_poll_step10](heap.assets/heap_poll_step10.png)
与元素入堆操作类似，**堆顶元素出堆操作的时间复杂度为 $O(\log n)$** 。 与元素入堆操作类似，**堆顶元素出堆操作的时间复杂度为 $O(\log n)$** 。
@ -862,7 +860,7 @@ $$
T(h) = 2^0h + 2^1(h-1) + 2^2(h-2) + \cdots + 2^{(h-1)}\times1 T(h) = 2^0h + 2^1(h-1) + 2^2(h-2) + \cdots + 2^{(h-1)}\times1
$$ $$
![heapify_count](heap.assets/heapify_count.png){ width="600" } ![heapify_count](heap.assets/heapify_count.png)
化简上式需要借助中学的数列知识，先对 $T(h)$ 乘以 $2$ ，易得 化简上式需要借助中学的数列知识，先对 $T(h)$ 乘以 $2$ ，易得