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docs/chapter_computational_complexity/exercises.md
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@@ -11,33 +11,287 @@ comments: true
下面两段代码都计算 $1 + 2 + \dots + n$（设 $n \ge 1$）。请把 `n` 设为 4，
按照程序实际执行的顺序回答问题，然后比较两种写法的效率。
```python
def sum_iter(n):
s = 0
for i in range(1, n + 1):
s += i
return s
=== "Python"
def sum_recur(n):
if n == 1:
return 1
return n + sum_recur(n - 1)
```
```python title="complexity_exercises.py"
def sum_iter(n: int) -> int:
"""迭代求和"""
res = 0
for i in range(1, n + 1):
res += i
return res
def sum_recur(n: int) -> int:
"""递归求和"""
if n == 1:
return 1
return n + sum_recur(n - 1)
```
=== "C++"
```cpp title="complexity_exercises.cpp"
/* 迭代求和 */
int sumIter(int n) {
int res = 0;
for (int i = 1; i <= n; ++i) {
res += i;
}
return res;
}
/* 递归求和 */
int sumRecur(int n) {
if (n == 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "Java"
```java title="complexity_exercises.java"
/* 迭代求和 */
int sumIter(int n) {
int res = 0;
for (int i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 递归求和 */
int sumRecur(int n) {
if (n == 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "C#"
```csharp title="complexity_exercises.cs"
/* 迭代求和 */
int SumIter(int n) {
int res = 0;
for (int i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 递归求和 */
int SumRecur(int n) {
if (n == 1) {
return 1;
}
return n + SumRecur(n - 1);
}
```
=== "Go"
```go title="complexity_exercises.go"
/* 迭代求和 */
func sumIter(n int) int {
res := 0
for i := 1; i <= n; i++ {
res += i
}
return res
}
/* 递归求和 */
func sumRecur(n int) int {
if n == 1 {
return 1
}
return n + sumRecur(n-1)
}
```
=== "Swift"
```swift title="complexity_exercises.swift"
/* 迭代求和 */
func sumIter(n: Int) -> Int {
var res = 0
for i in 1 ... n {
res += i
}
return res
}
/* 递归求和 */
func sumRecur(n: Int) -> Int {
if n == 1 {
return 1
}
return n + sumRecur(n: n - 1)
}
```
=== "JS"
```javascript title="complexity_exercises.js"
/* 迭代求和 */
function sumIter(n) {
let res = 0;
for (let i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 递归求和 */
function sumRecur(n) {
if (n === 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "TS"
```typescript title="complexity_exercises.ts"
/* 迭代求和 */
function sumIter(n: number): number {
let res = 0;
for (let i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 递归求和 */
function sumRecur(n: number): number {
if (n === 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "Dart"
```dart title="complexity_exercises.dart"
/* 迭代求和 */
int sumIter(int n) {
int res = 0;
for (int i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 递归求和 */
int sumRecur(int n) {
if (n == 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "Rust"
```rust title="complexity_exercises.rs"
/* 迭代求和 */
fn sum_iter(n: i32) -> i32 {
let mut res = 0;
for i in 1..=n {
res += i;
}
res
}
/* 递归求和 */
fn sum_recur(n: i32) -> i32 {
if n == 1 {
return 1;
}
n + sum_recur(n - 1)
}
```
=== "C"
```c title="complexity_exercises.c"
/* 迭代求和 */
int sumIter(int n) {
int res = 0;
for (int i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 递归求和 */
int sumRecur(int n) {
if (n == 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "Kotlin"
```kotlin title="complexity_exercises.kt"
/* 迭代求和 */
fun sumIter(n: Int): Int {
var res = 0
for (i in 1..n) {
res += i
}
return res
}
/* 递归求和 */
fun sumRecur(n: Int): Int {
if (n == 1) {
return 1
}
return n + sumRecur(n - 1)
}
```
=== "Ruby"
```ruby title="complexity_exercises.rb"
### 迭代求和 ###
def sum_iter(n)
res = 0
for i in 1..n
res += i
end
res
end
### 递归求和 ###
def sum_recur(n)
return 1 if n == 1
n + sum_recur(n - 1)
end
```
<!-- numbered-subquestions -->
1. 执行 `sum_iter(4)` 时，每轮循环结束后，变量 `s` 的值分别是多少？
2. 执行 `sum_recur(4)` 时，会依次调用哪些函数？从最深的一层开始返回时，结果怎样得到？
1. 输入 `n = 4` 执行迭代函数时，每轮循环结束后，累加变量 `res` 的值分别是多少？
2. 输入 `n = 4` 执行递归函数时，参数 `n` 会依次取哪些值？从最深的一层开始返回时，结果怎样得到？
3. 两种写法的时间复杂度和空间复杂度分别是多少？结合第 1、2 问的执行过程说明理由。
??? success "参考答案"
1. 循环变量 `i` 依次为 `1、2、3、4`，每轮结束后，`s` 依次变为
`1、3、6、10`，所以 `sum_iter(4)` 返回 10。
1. 循环变量 `i` 依次为 `1、2、3、4`，每轮结束后，`res` 依次变为
`1、3、6、10`，所以迭代函数返回 10。
2. 函数依次调用
`sum_recur(4) → sum_recur(3) → sum_recur(2) → sum_recur(1)`。
`sum_recur(1)` 返回 1，随后各层依次得到 `2 + 1 = 3`、`3 + 3 = 6`、`4 + 6 = 10`。
2. 参数 `n` 依次为 `4 → 3 → 2 → 1`。
最深一层返回 1，随后各层依次得到 `2 + 1 = 3`、`3 + 3 = 6`、`4 + 6 = 10`。
在最深处，4 次函数调用都尚未结束。
3. 两段代码都进行与 $n$ 成正比的循环或调用，因此时间复杂度均为 $O(n)$ 。
@@ -51,22 +305,414 @@ def sum_recur(n):
以下三个代码片段的输入均为正整数 $n$ 。请按时间复杂度从低到高排序，并写出各自的复杂度。
```python
# 片段一
s = 0
for i in range(n):
s += i
=== "Python"
# 片段二
s = 0
for i in range(n):
for j in range(i, n):
s += j
```python title="complexity_exercises.py"
def linear_loop(n: int) -> int:
"""线性阶循环"""
res = 0
for i in range(n):
res += i
return res
# 片段三
while n > 1:
n = n // 2
```
def quadratic_loop(n: int) -> int:
"""平方阶循环"""
res = 0
for i in range(n):
for j in range(i, n):
res += j
return res
def logarithmic_loop(n: int) -> int:
"""对数阶循环"""
while n > 1:
n //= 2
return n
```
=== "C++"
```cpp title="complexity_exercises.cpp"
/* 线性阶循环 */
int linearLoop(int n) {
int res = 0;
for (int i = 0; i < n; ++i) {
res += i;
}
return res;
}
/* 平方阶循环 */
int quadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; ++i) {
for (int j = i; j < n; ++j) {
res += j;
}
}
return res;
}
/* 对数阶循环 */
int logarithmicLoop(int n) {
while (n > 1) {
n /= 2;
}
return n;
}
```
=== "Java"
```java title="complexity_exercises.java"
/* 线性阶循环 */
int linearLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方阶循环 */
int quadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 对数阶循环 */
int logarithmicLoop(int n) {
while (n > 1) {
n /= 2;
}
return n;
}
```
=== "C#"
```csharp title="complexity_exercises.cs"
/* 线性阶循环 */
int LinearLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方阶循环 */
int QuadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 对数阶循环 */
int LogarithmicLoop(int n) {
while (n > 1) {
n /= 2;
}
return n;
}
```
=== "Go"
```go title="complexity_exercises.go"
/* 线性阶循环 */
func linearLoop(n int) int {
res := 0
for i := 0; i < n; i++ {
res += i
}
return res
}
/* 平方阶循环 */
func quadraticLoop(n int) int {
res := 0
for i := 0; i < n; i++ {
for j := i; j < n; j++ {
res += j
}
}
return res
}
/* 对数阶循环 */
func logarithmicLoop(n int) int {
for n > 1 {
n /= 2
}
return n
}
```
=== "Swift"
```swift title="complexity_exercises.swift"
/* 线性阶循环 */
func linearLoop(n: Int) -> Int {
var res = 0
for i in 0 ..< n {
res += i
}
return res
}
/* 平方阶循环 */
func quadraticLoop(n: Int) -> Int {
var res = 0
for i in 0 ..< n {
for j in i ..< n {
res += j
}
}
return res
}
/* 对数阶循环 */
func logarithmicLoop(n: Int) -> Int {
var n = n
while n > 1 {
n /= 2
}
return n
}
```
=== "JS"
```javascript title="complexity_exercises.js"
/* 线性阶循环 */
function linearLoop(n) {
let res = 0;
for (let i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方阶循环 */
function quadraticLoop(n) {
let res = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 对数阶循环 */
function logarithmicLoop(n) {
while (n > 1) {
n = Math.floor(n / 2);
}
return n;
}
```
=== "TS"
```typescript title="complexity_exercises.ts"
/* 线性阶循环 */
function linearLoop(n: number): number {
let res = 0;
for (let i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方阶循环 */
function quadraticLoop(n: number): number {
let res = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 对数阶循环 */
function logarithmicLoop(n: number): number {
while (n > 1) {
n = Math.floor(n / 2);
}
return n;
}
```
=== "Dart"
```dart title="complexity_exercises.dart"
/* 线性阶循环 */
int linearLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方阶循环 */
int quadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 对数阶循环 */
int logarithmicLoop(int n) {
while (n > 1) {
n ~/= 2;
}
return n;
}
```
=== "Rust"
```rust title="complexity_exercises.rs"
/* 线性阶循环 */
fn linear_loop(n: i32) -> i32 {
let mut res = 0;
for i in 0..n {
res += i;
}
res
}
/* 平方阶循环 */
fn quadratic_loop(n: i32) -> i32 {
let mut res = 0;
for i in 0..n {
for j in i..n {
res += j;
}
}
res
}
/* 对数阶循环 */
fn logarithmic_loop(mut n: i32) -> i32 {
while n > 1 {
n /= 2;
}
n
}
```
=== "C"
```c title="complexity_exercises.c"
/* 线性阶循环 */
int linearLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方阶循环 */
int quadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 对数阶循环 */
int logarithmicLoop(int n) {
while (n > 1) {
n /= 2;
}
return n;
}
```
=== "Kotlin"
```kotlin title="complexity_exercises.kt"
/* 线性阶循环 */
fun linearLoop(n: Int): Int {
var res = 0
for (i in 0 until n) {
res += i
}
return res
}
/* 平方阶循环 */
fun quadraticLoop(n: Int): Int {
var res = 0
for (i in 0 until n) {
for (j in i until n) {
res += j
}
}
return res
}
/* 对数阶循环 */
fun logarithmicLoop(n: Int): Int {
var value = n
while (value > 1) {
value /= 2
}
return value
}
```
=== "Ruby"
```ruby title="complexity_exercises.rb"
### 线性阶循环 ###
def linear_loop(n)
res = 0
for i in 0...n
res += i
end
res
end
### 平方阶循环 ###
def quadratic_loop(n)
res = 0
for i in 0...n
for j in i...n
res += j
end
end
res
end
### 对数阶循环 ###
def logarithmic_loop(n)
n /= 2 while n > 1
n
end
```
??? success "参考答案"