mirror of
https://github.com/krahets/hello-algo.git
synced 2024-12-25 01:36:29 +08:00
Compare commits
4 commits
554e45a4c8
...
2f7ffd2ae7
Author | SHA1 | Date | |
---|---|---|---|
|
2f7ffd2ae7 | ||
|
b6939da46c | ||
|
abf1f115bf | ||
|
96b551c4d0 |
14 changed files with 81 additions and 43 deletions
5
.gitignore
vendored
5
.gitignore
vendored
|
@ -1,7 +1,7 @@
|
||||||
# macOS
|
# macOS
|
||||||
.DS_Store
|
.DS_Store
|
||||||
|
|
||||||
# Editor
|
# editors
|
||||||
.vscode/
|
.vscode/
|
||||||
**/.idea
|
**/.idea
|
||||||
|
|
||||||
|
@ -12,6 +12,3 @@
|
||||||
/build
|
/build
|
||||||
/site
|
/site
|
||||||
/utils
|
/utils
|
||||||
|
|
||||||
# test script
|
|
||||||
test.sh
|
|
||||||
|
|
|
@ -53,6 +53,7 @@ int dequeIndex(ArrayDeque *deque, int i) {
|
||||||
return ((i + capacity(deque)) % capacity(deque));
|
return ((i + capacity(deque)) % capacity(deque));
|
||||||
}
|
}
|
||||||
|
|
||||||
|
|
||||||
/* 队首入队 */
|
/* 队首入队 */
|
||||||
void pushFirst(ArrayDeque *deque, int num) {
|
void pushFirst(ArrayDeque *deque, int num) {
|
||||||
if (deque->queSize == capacity(deque)) {
|
if (deque->queSize == capacity(deque)) {
|
||||||
|
@ -110,7 +111,17 @@ int popLast(ArrayDeque *deque) {
|
||||||
deque->queSize--;
|
deque->queSize--;
|
||||||
return num;
|
return num;
|
||||||
}
|
}
|
||||||
|
void print_deque(ArrayDeque *deque)
|
||||||
|
{
|
||||||
|
|
||||||
|
for(int i =0;i<deque->queSize-1;i++)
|
||||||
|
{
|
||||||
|
int pptr = (deque->front+i)%deque->queCapacity;
|
||||||
|
printf("%d,",deque->nums[pptr]);
|
||||||
|
}
|
||||||
|
//printf("\n");
|
||||||
|
printf("%d]\n",deque->nums[(deque->front+deque->queSize-1)%deque->queCapacity]);
|
||||||
|
}
|
||||||
/* Driver Code */
|
/* Driver Code */
|
||||||
int main() {
|
int main() {
|
||||||
/* 初始化队列 */
|
/* 初始化队列 */
|
||||||
|
@ -119,8 +130,8 @@ int main() {
|
||||||
pushLast(deque, 3);
|
pushLast(deque, 3);
|
||||||
pushLast(deque, 2);
|
pushLast(deque, 2);
|
||||||
pushLast(deque, 5);
|
pushLast(deque, 5);
|
||||||
printf("双向队列 deque = ");
|
printf("双向队列 deque = [");
|
||||||
printArray(deque->nums, deque->queSize);
|
print_deque(deque);
|
||||||
|
|
||||||
/* 访问元素 */
|
/* 访问元素 */
|
||||||
int peekFirstNum = peekFirst(deque);
|
int peekFirstNum = peekFirst(deque);
|
||||||
|
@ -130,19 +141,19 @@ int main() {
|
||||||
|
|
||||||
/* 元素入队 */
|
/* 元素入队 */
|
||||||
pushLast(deque, 4);
|
pushLast(deque, 4);
|
||||||
printf("元素 4 队尾入队后 deque = ");
|
printf("元素 4 队尾入队后 deque = [");
|
||||||
printArray(deque->nums, deque->queSize);
|
print_deque(deque);
|
||||||
pushFirst(deque, 1);
|
pushFirst(deque, 1);
|
||||||
printf("元素 1 队首入队后 deque = ");
|
printf("元素 1 队首入队后 deque = [");
|
||||||
printArray(deque->nums, deque->queSize);
|
print_deque(deque);
|
||||||
|
|
||||||
/* 元素出队 */
|
/* 元素出队 */
|
||||||
int popLastNum = popLast(deque);
|
int popLastNum = popLast(deque);
|
||||||
printf("队尾出队元素 = %d ,队尾出队后 deque= ", popLastNum);
|
printf("队尾出队元素 = %d ,队尾出队后 deque= [", popLastNum);
|
||||||
printArray(deque->nums, deque->queSize);
|
print_deque(deque);
|
||||||
int popFirstNum = popFirst(deque);
|
int popFirstNum = popFirst(deque);
|
||||||
printf("队首出队元素 = %d ,队首出队后 deque= ", popFirstNum);
|
printf("队首出队元素 = %d ,队首出队后 deque= [", popFirstNum);
|
||||||
printArray(deque->nums, deque->queSize);
|
print_deque(deque);
|
||||||
|
|
||||||
/* 获取队列的长度 */
|
/* 获取队列的长度 */
|
||||||
int dequeSize = size(deque);
|
int dequeSize = size(deque);
|
||||||
|
|
|
@ -73,7 +73,17 @@ int pop(ArrayQueue *queue) {
|
||||||
queue->queSize--;
|
queue->queSize--;
|
||||||
return num;
|
return num;
|
||||||
}
|
}
|
||||||
|
void print_queue(ArrayQueue *queue)
|
||||||
|
{
|
||||||
|
|
||||||
|
for(int i =0;i<queue->queSize-1;i++)
|
||||||
|
{
|
||||||
|
int pptr = (queue->front+i)%queue->queCapacity;
|
||||||
|
printf("%d,",queue->nums[pptr]);
|
||||||
|
}
|
||||||
|
//printf("\n");
|
||||||
|
printf("%d]\n",queue->nums[(queue->front+queue->queSize-1)%queue->queCapacity]);
|
||||||
|
}
|
||||||
/* Driver Code */
|
/* Driver Code */
|
||||||
int main() {
|
int main() {
|
||||||
/* 初始化队列 */
|
/* 初始化队列 */
|
||||||
|
@ -95,8 +105,8 @@ int main() {
|
||||||
|
|
||||||
/* 元素出队 */
|
/* 元素出队 */
|
||||||
peekNum = pop(queue);
|
peekNum = pop(queue);
|
||||||
printf("出队元素 pop = %d ,出队后 queue = ", peekNum);
|
printf("出队元素 pop = %d ,出队后 queue = [", peekNum);
|
||||||
printArray(queue->nums, queue->queSize);
|
print_queue(queue);
|
||||||
|
|
||||||
/* 获取队列的长度 */
|
/* 获取队列的长度 */
|
||||||
int queueSize = size(queue);
|
int queueSize = size(queue);
|
||||||
|
@ -111,7 +121,7 @@ int main() {
|
||||||
push(queue, i);
|
push(queue, i);
|
||||||
pop(queue);
|
pop(queue);
|
||||||
printf("第 %d 轮入队 + 出队后 queue = ", i);
|
printf("第 %d 轮入队 + 出队后 queue = ", i);
|
||||||
printArray(queue->nums, queue->queSize);
|
print_queue(queue);
|
||||||
}
|
}
|
||||||
|
|
||||||
// 释放内存
|
// 释放内存
|
||||||
|
|
|
@ -71,6 +71,16 @@
|
||||||
|
|
||||||
另一方面,必要使用链表的情况主要是二叉树和图。栈和队列往往会使用编程语言提供的 `stack` 和 `queue` ,而非链表。
|
另一方面,必要使用链表的情况主要是二叉树和图。栈和队列往往会使用编程语言提供的 `stack` 和 `queue` ,而非链表。
|
||||||
|
|
||||||
**Q**:初始化列表 `res = [0] * self.size()` 操作,会导致 `res` 的每个元素引用相同的地址吗?
|
**Q**:操作 `res = [[0]] * n` 生成了一个二维列表,其中每一个 `[0]` 都是独立的吗?
|
||||||
|
|
||||||
不会。但二维数组会有这个问题,例如初始化二维列表 `res = [[0]] * self.size()` ,则多次引用了同一个列表 `[0]` 。
|
不是独立的。此二维列表中,所有的 `[0]` 实际上是同一个对象的引用。如果我们修改其中一个元素,会发现所有的对应元素都会随之改变。
|
||||||
|
|
||||||
|
如果希望二维列表中的每个 `[0]` 都是独立的,可以使用 `res = [[0] for _ in range(n)]` 来实现。这种方式的原理是初始化了 $n$ 个独立的 `[0]` 列表对象。
|
||||||
|
|
||||||
|
**Q**:操作 `res = [0] * n` 生成了一个列表,其中每一个整数 0 都是独立的吗?
|
||||||
|
|
||||||
|
在该列表中,所有整数 0 都是同一个对象的引用。这是因为 Python 对小整数(通常是 -5 到 256)采用了缓存池机制,以便最大化对象复用,从而提升性能。
|
||||||
|
|
||||||
|
虽然它们指向同一个对象,但我们仍然可以独立修改列表中的每个元素,这是因为 Python 的整数是“不可变对象”。当我们修改某个元素时,实际上是切换为另一个对象的引用,而不是改变原有对象本身。
|
||||||
|
|
||||||
|
然而,当列表元素是“可变对象”时(例如列表、字典或类实例等),修改某个元素会直接改变该对象本身,所有引用该对象的元素都会产生相同变化。
|
||||||
|
|
|
@ -5,7 +5,7 @@
|
||||||
- 整理扑克的过程与插入排序算法非常类似。插入排序算法适合排序小型数据集。
|
- 整理扑克的过程与插入排序算法非常类似。插入排序算法适合排序小型数据集。
|
||||||
- 货币找零的步骤本质上是贪心算法,每一步都采取当前看来最好的选择。
|
- 货币找零的步骤本质上是贪心算法,每一步都采取当前看来最好的选择。
|
||||||
- 算法是在有限时间内解决特定问题的一组指令或操作步骤,而数据结构是计算机中组织和存储数据的方式。
|
- 算法是在有限时间内解决特定问题的一组指令或操作步骤,而数据结构是计算机中组织和存储数据的方式。
|
||||||
- 数据结构与算法紧密相连。数据结构是算法的基石,而算法是数据结构发挥作用的舞台。
|
- 数据结构与算法紧密相连。数据结构是算法的基石,而算法为数据结构注入生命力。
|
||||||
- 我们可以将数据结构与算法类比为拼装积木,积木代表数据,积木的形状和连接方式等代表数据结构,拼装积木的步骤则对应算法。
|
- 我们可以将数据结构与算法类比为拼装积木,积木代表数据,积木的形状和连接方式等代表数据结构,拼装积木的步骤则对应算法。
|
||||||
|
|
||||||
### Q & A
|
### Q & A
|
||||||
|
|
|
@ -26,7 +26,7 @@
|
||||||
如下图所示,数据结构与算法高度相关、紧密结合,具体表现在以下三个方面。
|
如下图所示,数据结构与算法高度相关、紧密结合,具体表现在以下三个方面。
|
||||||
|
|
||||||
- 数据结构是算法的基石。数据结构为算法提供了结构化存储的数据,以及操作数据的方法。
|
- 数据结构是算法的基石。数据结构为算法提供了结构化存储的数据,以及操作数据的方法。
|
||||||
- 算法是数据结构发挥作用的舞台。数据结构本身仅存储数据信息,结合算法才能解决特定问题。
|
- 算法为数据结构注入生命力。数据结构本身仅存储数据信息,结合算法才能解决特定问题。
|
||||||
- 算法通常可以基于不同的数据结构实现,但执行效率可能相差很大,选择合适的数据结构是关键。
|
- 算法通常可以基于不同的数据结构实现,但执行效率可能相差很大,选择合适的数据结构是关键。
|
||||||
|
|
||||||
![数据结构与算法的关系](what_is_dsa.assets/relationship_between_data_structure_and_algorithm.png)
|
![数据结构与算法的关系](what_is_dsa.assets/relationship_between_data_structure_and_algorithm.png)
|
||||||
|
|
|
@ -32,14 +32,14 @@ That is, our contributors are computer scientists, engineers, and students from
|
||||||
> [!important]
|
> [!important]
|
||||||
> Before diving in, ensure you're comfortable with the GitHub pull request workflow and have read the "Translation standards" and "Pseudo-code for translation" below.
|
> Before diving in, ensure you're comfortable with the GitHub pull request workflow and have read the "Translation standards" and "Pseudo-code for translation" below.
|
||||||
|
|
||||||
1. **Self-assignment**: Visit [GitHub projects](https://github.com/users/krahets/projects/2/views/4) to select an unclaimed task and mark it as `In Progress`.
|
1. **Task assignment**: Self-assign a task in the Notion workspace.
|
||||||
2. **Translation**: We encourage preserving the original meaning while ensuring the translation is natural and fluent.
|
2. **Translation**: Optimize the translation on your local PC, referring to the “Translation Pseudo-Code” section below for more details.
|
||||||
3. **Peer review**: Please carefully check your changes before submitting a Pull Request (PR). After approval by two reviewers, it will be merged into the project.
|
3. **Peer review**: Carefully review your changes before submitting a Pull Request (PR). The PR will be merged into the main branch after approval from two reviewers.
|
||||||
|
|
||||||
## Translation standards
|
## Translation standards
|
||||||
|
|
||||||
> [!tip]
|
> [!tip]
|
||||||
> The "Accuracy" and "Authenticity" are primarily handled by native Chinese speakers and native English speakers, respectively.
|
> **The "Accuracy" and "Authenticity" are primarily handled by native Chinese speakers and native English speakers, respectively.**
|
||||||
>
|
>
|
||||||
> In some instances, "Accuracy (consistency)" and "Authenticity" represent a trade-off, where optimizing one aspect could significantly affect the other. In such cases, please leave a comment in the pull request for discussion.
|
> In some instances, "Accuracy (consistency)" and "Authenticity" represent a trade-off, where optimizing one aspect could significantly affect the other. In such cases, please leave a comment in the pull request for discussion.
|
||||||
|
|
||||||
|
|
|
@ -1,6 +1,6 @@
|
||||||
# Hash optimization strategies
|
# Hash optimization strategies
|
||||||
|
|
||||||
In algorithm problems, **we often reduce the time complexity of algorithms by replacing linear search with hash search**. Let's use an algorithm problem to deepen understanding.
|
In algorithm problems, **we often reduce the time complexity of an algorithm by replacing a linear search with a hash-based search**. Let's use an algorithm problem to deepen the understanding.
|
||||||
|
|
||||||
!!! question
|
!!! question
|
||||||
|
|
||||||
|
@ -8,7 +8,7 @@ In algorithm problems, **we often reduce the time complexity of algorithms by re
|
||||||
|
|
||||||
## Linear search: trading time for space
|
## Linear search: trading time for space
|
||||||
|
|
||||||
Consider traversing all possible combinations directly. As shown in the figure below, we initiate a two-layer loop, and in each round, we determine whether the sum of the two integers equals `target`. If so, we return their indices.
|
Consider traversing through all possible combinations directly. As shown in the figure below, we initiate a nested loop, and in each iteration, we determine whether the sum of the two integers equals `target`. If so, we return their indices.
|
||||||
|
|
||||||
![Linear search solution for two-sum problem](replace_linear_by_hashing.assets/two_sum_brute_force.png)
|
![Linear search solution for two-sum problem](replace_linear_by_hashing.assets/two_sum_brute_force.png)
|
||||||
|
|
||||||
|
@ -18,11 +18,11 @@ The code is shown below:
|
||||||
[file]{two_sum}-[class]{}-[func]{two_sum_brute_force}
|
[file]{two_sum}-[class]{}-[func]{two_sum_brute_force}
|
||||||
```
|
```
|
||||||
|
|
||||||
This method has a time complexity of $O(n^2)$ and a space complexity of $O(1)$, which is very time-consuming with large data volumes.
|
This method has a time complexity of $O(n^2)$ and a space complexity of $O(1)$, which can be very time-consuming with large data volumes.
|
||||||
|
|
||||||
## Hash search: trading space for time
|
## Hash search: trading space for time
|
||||||
|
|
||||||
Consider using a hash table, with key-value pairs being the array elements and their indices, respectively. Loop through the array, performing the steps shown in the figure below each round.
|
Consider using a hash table, where the key-value pairs are the array elements and their indices, respectively. Loop through the array, performing the steps shown in the figure below during each iteration.
|
||||||
|
|
||||||
1. Check if the number `target - nums[i]` is in the hash table. If so, directly return the indices of these two elements.
|
1. Check if the number `target - nums[i]` is in the hash table. If so, directly return the indices of these two elements.
|
||||||
2. Add the key-value pair `nums[i]` and index `i` to the hash table.
|
2. Add the key-value pair `nums[i]` and index `i` to the hash table.
|
||||||
|
@ -42,6 +42,6 @@ The implementation code is shown below, requiring only a single loop:
|
||||||
[file]{two_sum}-[class]{}-[func]{two_sum_hash_table}
|
[file]{two_sum}-[class]{}-[func]{two_sum_hash_table}
|
||||||
```
|
```
|
||||||
|
|
||||||
This method reduces the time complexity from $O(n^2)$ to $O(n)$ by using hash search, greatly improving the running efficiency.
|
This method reduces the time complexity from $O(n^2)$ to $O(n)$ by using hash search, significantly enhancing runtime efficiency.
|
||||||
|
|
||||||
As it requires maintaining an additional hash table, the space complexity is $O(n)$. **Nevertheless, this method has a more balanced time-space efficiency overall, making it the optimal solution for this problem**.
|
As it requires maintaining an additional hash table, the space complexity is $O(n)$. **Nevertheless, this method has a more balanced time-space efficiency overall, making it the optimal solution for this problem**.
|
||||||
|
|
|
@ -414,7 +414,7 @@
|
||||||
<!-- contributors -->
|
<!-- contributors -->
|
||||||
<div style="margin: 2em auto;">
|
<div style="margin: 2em auto;">
|
||||||
<h3>Contributors</h3>
|
<h3>Contributors</h3>
|
||||||
<p>This book has been optimized by the efforts of over 180 contributors. We sincerely thank them for their invaluable time and contributions!</p>
|
<p>This book has been refined by the efforts of over 180 contributors. We sincerely thank them for their invaluable time and contributions!</p>
|
||||||
<a href="https://github.com/krahets/hello-algo/graphs/contributors">
|
<a href="https://github.com/krahets/hello-algo/graphs/contributors">
|
||||||
<img src="https://contrib.rocks/image?repo=krahets/hello-algo&max=300&columns=16" alt="Contributors" style="width: 100%; max-width: 38.5em;">
|
<img src="https://contrib.rocks/image?repo=krahets/hello-algo&max=300&columns=16" alt="Contributors" style="width: 100%; max-width: 38.5em;">
|
||||||
</a>
|
</a>
|
||||||
|
|
|
@ -115,9 +115,9 @@ int main() {
|
||||||
int i = 1;
|
int i = 1;
|
||||||
int l = abt.left(i), r = abt.right(i), p = abt.parent(i);
|
int l = abt.left(i), r = abt.right(i), p = abt.parent(i);
|
||||||
cout << "\n當前節點的索引為 " << i << ",值為 " << abt.val(i) << "\n";
|
cout << "\n當前節點的索引為 " << i << ",值為 " << abt.val(i) << "\n";
|
||||||
cout << "其左子節點的索引為 " << l << ",值為 " << (l != INT_MAX ? to_string(abt.val(l)) : "nullptr") << "\n";
|
cout << "其左子節點的索引為 " << l << ",值為 " << (abt.val(l) != INT_MAX ? to_string(abt.val(l)) : "nullptr") << "\n";
|
||||||
cout << "其右子節點的索引為 " << r << ",值為 " << (r != INT_MAX ? to_string(abt.val(r)) : "nullptr") << "\n";
|
cout << "其右子節點的索引為 " << r << ",值為 " << (abt.val(r) != INT_MAX ? to_string(abt.val(r)) : "nullptr") << "\n";
|
||||||
cout << "其父節點的索引為 " << p << ",值為 " << (p != INT_MAX ? to_string(abt.val(p)) : "nullptr") << "\n";
|
cout << "其父節點的索引為 " << p << ",值為 " << (abt.val(p) != INT_MAX ? to_string(abt.val(p)) : "nullptr") << "\n";
|
||||||
|
|
||||||
// 走訪樹
|
// 走訪樹
|
||||||
vector<int> res = abt.levelOrder();
|
vector<int> res = abt.levelOrder();
|
||||||
|
|
|
@ -71,6 +71,16 @@
|
||||||
|
|
||||||
另一方面,必要使用鏈結串列的情況主要是二元樹和圖。堆疊和佇列往往會使用程式語言提供的 `stack` 和 `queue` ,而非鏈結串列。
|
另一方面,必要使用鏈結串列的情況主要是二元樹和圖。堆疊和佇列往往會使用程式語言提供的 `stack` 和 `queue` ,而非鏈結串列。
|
||||||
|
|
||||||
**Q**:初始化串列 `res = [0] * self.size()` 操作,會導致 `res` 的每個元素引用相同的位址嗎?
|
**Q**:操作 `res = [[0]] * n` 生成了一個二維串列,其中每一個 `[0]` 都是獨立的嗎?
|
||||||
|
|
||||||
不會。但二維陣列會有這個問題,例如初始化二維串列 `res = [[0]] * self.size()` ,則多次引用了同一個串列 `[0]` 。
|
不是獨立的。此二維串列中,所有的 `[0]` 實際上是同一個物件的引用。如果我們修改其中一個元素,會發現所有的對應元素都會隨之改變。
|
||||||
|
|
||||||
|
如果希望二維串列中的每個 `[0]` 都是獨立的,可以使用 `res = [[0] for _ in range(n)]` 來實現。這種方式的原理是初始化了 $n$ 個獨立的 `[0]` 串列物件。
|
||||||
|
|
||||||
|
**Q**:操作 `res = [0] * n` 生成了一個串列,其中每一個整數 0 都是獨立的嗎?
|
||||||
|
|
||||||
|
在該串列中,所有整數 0 都是同一個物件的引用。這是因為 Python 對小整數(通常是 -5 到 256)採用了快取池機制,以便最大化物件複用,從而提升效能。
|
||||||
|
|
||||||
|
雖然它們指向同一個物件,但我們仍然可以獨立修改串列中的每個元素,這是因為 Python 的整數是“不可變物件”。當我們修改某個元素時,實際上是切換為另一個物件的引用,而不是改變原有物件本身。
|
||||||
|
|
||||||
|
然而,當串列元素是“可變物件”時(例如串列、字典或類別例項等),修改某個元素會直接改變該物件本身,所有引用該物件的元素都會產生相同變化。
|
||||||
|
|
|
@ -1120,7 +1120,7 @@ $$
|
||||||
|
|
||||||
生物學的“細胞分裂”是指數階增長的典型例子:初始狀態為 $1$ 個細胞,分裂一輪後變為 $2$ 個,分裂兩輪後變為 $4$ 個,以此類推,分裂 $n$ 輪後有 $2^n$ 個細胞。
|
生物學的“細胞分裂”是指數階增長的典型例子:初始狀態為 $1$ 個細胞,分裂一輪後變為 $2$ 個,分裂兩輪後變為 $4$ 個,以此類推,分裂 $n$ 輪後有 $2^n$ 個細胞。
|
||||||
|
|
||||||
下圖和以下程式碼模擬了細胞分裂的過程,時間複雜度為 $O(2^n)$ :
|
下圖和以下程式碼模擬了細胞分裂的過程,時間複雜度為 $O(2^n)$ 。請注意,輸入 $n$ 表示分裂輪數,返回值 `count` 表示總分裂次數。
|
||||||
|
|
||||||
```src
|
```src
|
||||||
[file]{time_complexity}-[class]{}-[func]{exponential}
|
[file]{time_complexity}-[class]{}-[func]{exponential}
|
||||||
|
|
|
@ -5,7 +5,7 @@
|
||||||
- 整理撲克的過程與插入排序演算法非常類似。插入排序演算法適合排序小型資料集。
|
- 整理撲克的過程與插入排序演算法非常類似。插入排序演算法適合排序小型資料集。
|
||||||
- 貨幣找零的步驟本質上是貪婪演算法,每一步都採取當前看來最好的選擇。
|
- 貨幣找零的步驟本質上是貪婪演算法,每一步都採取當前看來最好的選擇。
|
||||||
- 演算法是在有限時間內解決特定問題的一組指令或操作步驟,而資料結構是計算機中組織和儲存資料的方式。
|
- 演算法是在有限時間內解決特定問題的一組指令或操作步驟,而資料結構是計算機中組織和儲存資料的方式。
|
||||||
- 資料結構與演算法緊密相連。資料結構是演算法的基石,而演算法是資料結構發揮作用的舞臺。
|
- 資料結構與演算法緊密相連。資料結構是演算法的基石,而演算法為資料結構注入生命力。
|
||||||
- 我們可以將資料結構與演算法類比為拼裝積木,積木代表資料,積木的形狀和連線方式等代表資料結構,拼裝積木的步驟則對應演算法。
|
- 我們可以將資料結構與演算法類比為拼裝積木,積木代表資料,積木的形狀和連線方式等代表資料結構,拼裝積木的步驟則對應演算法。
|
||||||
|
|
||||||
### Q & A
|
### Q & A
|
||||||
|
|
|
@ -26,7 +26,7 @@
|
||||||
如下圖所示,資料結構與演算法高度相關、緊密結合,具體表現在以下三個方面。
|
如下圖所示,資料結構與演算法高度相關、緊密結合,具體表現在以下三個方面。
|
||||||
|
|
||||||
- 資料結構是演算法的基石。資料結構為演算法提供了結構化儲存的資料,以及操作資料的方法。
|
- 資料結構是演算法的基石。資料結構為演算法提供了結構化儲存的資料,以及操作資料的方法。
|
||||||
- 演算法是資料結構發揮作用的舞臺。資料結構本身僅儲存資料資訊,結合演算法才能解決特定問題。
|
- 演算法為資料結構注入生命力。資料結構本身僅儲存資料資訊,結合演算法才能解決特定問題。
|
||||||
- 演算法通常可以基於不同的資料結構實現,但執行效率可能相差很大,選擇合適的資料結構是關鍵。
|
- 演算法通常可以基於不同的資料結構實現,但執行效率可能相差很大,選擇合適的資料結構是關鍵。
|
||||||
|
|
||||||
![資料結構與演算法的關係](what_is_dsa.assets/relationship_between_data_structure_and_algorithm.png)
|
![資料結構與演算法的關係](what_is_dsa.assets/relationship_between_data_structure_and_algorithm.png)
|
||||||
|
|
Loading…
Reference in a new issue