diff --git a/docs/chapter_heap/heap.md b/docs/chapter_heap/heap.md
index d674b3ba0..92ba6d1e6 100644
--- a/docs/chapter_heap/heap.md
+++ b/docs/chapter_heap/heap.md
@@ -10,7 +10,7 @@ comments: true
- 「小顶堆 Min Heap」,任意结点的值 $\leq$ 其子结点的值;
## 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。
-
+![representation_of_heap](heap.assets/representation_of_heap.png)
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 2"
- ![heap_push_step2](heap.assets/heap_push_step2.png){ width="600" }
+ ![heap_push_step2](heap.assets/heap_push_step2.png)
=== "Step 3"
- ![heap_push_step3](heap.assets/heap_push_step3.png){ width="600" }
+ ![heap_push_step3](heap.assets/heap_push_step3.png)
=== "Step 4"
- ![heap_push_step4](heap.assets/heap_push_step4.png){ width="600" }
+ ![heap_push_step4](heap.assets/heap_push_step4.png)
=== "Step 5"
- ![heap_push_step5](heap.assets/heap_push_step5.png){ width="600" }
+ ![heap_push_step5](heap.assets/heap_push_step5.png)
=== "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 $O(\log n)$** 。
@@ -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。
=== "Step 1"
- ![heap_poll_step1](heap.assets/heap_poll_step1.png){ width="600" }
+ ![heap_poll_step1](heap.assets/heap_poll_step1.png)
=== "Step 2"
- ![heap_poll_step2](heap.assets/heap_poll_step2.png){ width="600" }
+ ![heap_poll_step2](heap.assets/heap_poll_step2.png)
=== "Step 3"
- ![heap_poll_step3](heap.assets/heap_poll_step3.png){ width="600" }
+ ![heap_poll_step3](heap.assets/heap_poll_step3.png)
=== "Step 4"
- ![heap_poll_step4](heap.assets/heap_poll_step4.png){ width="600" }
+ ![heap_poll_step4](heap.assets/heap_poll_step4.png)
=== "Step 5"
- ![heap_poll_step5](heap.assets/heap_poll_step5.png){ width="600" }
+ ![heap_poll_step5](heap.assets/heap_poll_step5.png)
=== "Step 6"
- ![heap_poll_step6](heap.assets/heap_poll_step6.png){ width="600" }
+ ![heap_poll_step6](heap.assets/heap_poll_step6.png)
=== "Step 7"
- ![heap_poll_step7](heap.assets/heap_poll_step7.png){ width="600" }
+ ![heap_poll_step7](heap.assets/heap_poll_step7.png)
=== "Step 8"
- ![heap_poll_step8](heap.assets/heap_poll_step8.png){ width="600" }
+ ![heap_poll_step8](heap.assets/heap_poll_step8.png)
=== "Step 9"
- ![heap_poll_step9](heap.assets/heap_poll_step9.png){ width="600" }
+ ![heap_poll_step9](heap.assets/heap_poll_step9.png)
=== "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 $O(\log n)$** 。
@@ -862,7 +860,7 @@ $$
T(h) = 2^0h + 2^1(h-1) + 2^2(h-2) + \cdots + 2^{(h-1)}\times1
$$
-![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