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* copy zig codes of chapter_array_and_linkedlist and chapter_computational_complexity to markdown files

* Update time_complexity.md

---------

Co-authored-by: Yudong Jin <krahets@163.com>
This commit is contained in:
sjinzh
2023-02-03 19:15:34 +08:00
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89
docs/chapter_computational_complexity/space_complexity.md
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@ -643,7 +643,29 @@ $$
=== "Zig"
```zig title="space_complexity.zig"
// 常数阶
fn constant(n: i32) void {
// 常量、变量、对象占用 O(1) 空间
const a: i32 = 0;
var b: i32 = 0;
var nums = [_]i32{0}**10000;
var node = inc.ListNode(i32){.val = 0};
var i: i32 = 0;
// 循环中的变量占用 O(1) 空间
while (i < n) : (i += 1) {
var c: i32 = 0;
_ = c;
}
// 循环中的函数占用 O(1) 空间
i = 0;
while (i < n) : (i += 1) {
_ = function();
}
_ = a;
_ = b;
_ = nums;
_ = node;
}
```
### 线性阶 $O(n)$
@ -781,7 +803,28 @@ $$
=== "Zig"
```zig title="space_complexity.zig"
// 线性阶
fn linear(comptime n: i32) !void {
// 长度为 n 的数组占用 O(n) 空间
var nums = [_]i32{0}**n;
// 长度为 n 的列表占用 O(n) 空间
var nodes = std.ArrayList(i32).init(std.heap.page_allocator);
defer nodes.deinit();
var i: i32 = 0;
while (i < n) : (i += 1) {
try nodes.append(i);
}
// 长度为 n 的哈希表占用 O(n) 空间
var map = std.AutoArrayHashMap(i32, []const u8).init(std.heap.page_allocator);
defer map.deinit();
var j: i32 = 0;
while (j < n) : (j += 1) {
const string = try std.fmt.allocPrint(std.heap.page_allocator, "{d}", .{j});
defer std.heap.page_allocator.free(string);
try map.put(i, string);
}
_ = nums;
}
```
以下递归函数会同时存在 $n$ 个未返回的 `algorithm()` 函数，使用 $O(n)$ 大小的栈帧空间。
@ -877,7 +920,12 @@ $$
=== "Zig"
```zig title="space_complexity.zig"
// 线性阶（递归实现）
fn linearRecur(comptime n: i32) void {
std.debug.print("递归 n = {}\n", .{n});
if (n == 1) return;
linearRecur(n - 1);
}
```
![space_complexity_recursive_linear](space_complexity.assets/space_complexity_recursive_linear.png)
@ -1000,7 +1048,22 @@ $$
=== "Zig"
```zig title="space_complexity.zig"
// 平方阶
fn quadratic(n: i32) !void {
// 二维列表占用 O(n^2) 空间
var nodes = std.ArrayList(std.ArrayList(i32)).init(std.heap.page_allocator);
defer nodes.deinit();
var i: i32 = 0;
while (i < n) : (i += 1) {
var tmp = std.ArrayList(i32).init(std.heap.page_allocator);
defer tmp.deinit();
var j: i32 = 0;
while (j < n) : (j += 1) {
try tmp.append(0);
}
try nodes.append(tmp);
}
}
```
在以下递归函数中，同时存在 $n$ 个未返回的 `algorithm()` ，并且每个函数中都初始化了一个数组，长度分别为 $n, n-1, n-2, ..., 2, 1$ ，平均长度为 $\frac{n}{2}$ ，因此总体使用 $O(n^2)$ 空间。
@ -1103,7 +1166,13 @@ $$
=== "Zig"
```zig title="space_complexity.zig"
// 平方阶（递归实现）
fn quadraticRecur(comptime n: i32) i32 {
if (n <= 0) return 0;
var nums = [_]i32{0}**n;
std.debug.print("递归 n = {} 中的 nums 长度 = {}\n", .{n, nums.len});
return quadraticRecur(n - 1);
}
```
![space_complexity_recursive_quadratic](space_complexity.assets/space_complexity_recursive_quadratic.png)
@ -1217,7 +1286,15 @@ $$
=== "Zig"
```zig title="space_complexity.zig"
// 指数阶（建立满二叉树）
fn buildTree(mem_allocator: std.mem.Allocator, n: i32) !?*inc.TreeNode(i32) {
if (n == 0) return null;
const root = try mem_allocator.create(inc.TreeNode(i32));
root.init(0);
root.left = try buildTree(mem_allocator, n - 1);
root.right = try buildTree(mem_allocator, n - 1);
return root;
}
```
![space_complexity_exponential](space_complexity.assets/space_complexity_exponential.png)

37
docs/chapter_computational_complexity/space_time_tradeoff.md
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@ -178,7 +178,23 @@ comments: true
=== "Zig"
```zig title="leetcode_two_sum.zig"
const SolutionBruteForce = struct {
pub fn twoSum(self: *SolutionBruteForce, nums: []i32, target: i32) [2]i32 {
_ = self;
var size: usize = nums.len;
var i: usize = 0;
// 两层循环，时间复杂度 O(n^2)
while (i < size - 1) : (i += 1) {
var j = i + 1;
while (j < size) : (j += 1) {
if (nums[i] + nums[j] == target) {
return [_]i32{@intCast(i32, i), @intCast(i32, j)};
}
}
}
return undefined;
}
};
```
### 方法二：辅助哈希表
@ -347,5 +363,22 @@ comments: true
=== "Zig"
```zig title="leetcode_two_sum.zig"
const SolutionHashMap = struct {
pub fn twoSum(self: *SolutionHashMap, nums: []i32, target: i32) ![2]i32 {
_ = self;
var size: usize = nums.len;
// 辅助哈希表，空间复杂度 O(n)
var dic = std.AutoHashMap(i32, i32).init(std.heap.page_allocator);
defer dic.deinit();
var i: usize = 0;
// 单层循环，时间复杂度 O(n)
while (i < size) : (i += 1) {
if (dic.contains(target - nums[i])) {
return [_]i32{dic.get(target - nums[i]).?, @intCast(i32, i)};
}
try dic.put(nums[i], @intCast(i32, i));
}
return undefined;
}
};
```

156
docs/chapter_computational_complexity/time_complexity.md
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@ -914,7 +914,17 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 常数阶
fn constant(n: i32) i32 {
_ = n;
var count: i32 = 0;
const size: i32 = 100_000;
var i: i32 = 0;
while(i<size) : (i += 1) {
count += 1;
}
return count;
}
```
### 线性阶 $O(n)$
@ -1033,7 +1043,15 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 线性阶
fn linear(n: i32) i32 {
var count: i32 = 0;
var i: i32 = 0;
while (i < n) : (i += 1) {
count += 1;
}
return count;
}
```
「遍历数组」和「遍历链表」等操作，时间复杂度都为 $O(n)$ ，其中 $n$ 为数组或链表的长度。
@ -1171,7 +1189,15 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 线性阶（遍历数组）
fn arrayTraversal(nums: []i32) i32 {
var count: i32 = 0;
// 循环次数与数组长度成正比
for (nums) |_| {
count += 1;
}
return count;
}
```
### 平方阶 $O(n^2)$
@ -1325,7 +1351,19 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 平方阶
fn quadratic(n: i32) i32 {
var count: i32 = 0;
var i: i32 = 0;
// 循环次数与数组长度成平方关系
while (i < n) : (i += 1) {
var j: i32 = 0;
while (j < n) : (j += 1) {
count += 1;
}
}
return count;
}
```
![time_complexity_constant_linear_quadratic](time_complexity.assets/time_complexity_constant_linear_quadratic.png)
@ -1551,7 +1589,26 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 平方阶（冒泡排序）
fn bubbleSort(nums: []i32) i32 {
var count: i32 = 0; // 计数器
// 外循环：待排序元素数量为 n-1, n-2, ..., 1
var i: i32 = @intCast(i32, nums.len ) - 1;
while (i > 0) : (i -= 1) {
var j: usize = 0;
// 内循环：冒泡操作
while (j < i) : (j += 1) {
if (nums[j] > nums[j + 1]) {
// 交换 nums[j] 与 nums[j + 1]
var tmp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = tmp;
count += 3; // 元素交换包含 3 个单元操作
}
}
}
return count;
}
```
### 指数阶 $O(2^n)$
@ -1732,7 +1789,22 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 指数阶（循环实现）
fn exponential(n: i32) i32{
var count: i32 = 0;
var bas: i32 = 1;
var i: i32 = 0;
// cell 每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
while (i < n) : (i += 1) {
var j: i32 = 0;
while (j < bas) : (j += 1) {
count += 1;
}
bas *= 2;
}
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
return count;
}
```
![time_complexity_exponential](time_complexity.assets/time_complexity_exponential.png)
@ -1839,7 +1911,11 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 指数阶（递归实现）
fn expRecur(n: i32) i32{
if (n == 1) return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
```
### 对数阶 $O(\log n)$
@ -1980,7 +2056,18 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 对数阶（循环实现）
fn logarithmic(n: f32) i32
{
var count: i32 = 0;
var n_var = n;
while (n_var > 1)
{
n_var = n_var / 2;
count +=1;
}
return count;
}
```
![time_complexity_logarithmic](time_complexity.assets/time_complexity_logarithmic.png)
@ -2086,7 +2173,12 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 对数阶（递归实现）
fn logRecur(n: f32) i32
{
if (n <= 1) return 0;
return logRecur(n / 2) + 1;
}
```
### 线性对数阶 $O(n \log n)$
@ -2234,7 +2326,18 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 线性对数阶
fn linearLogRecur(n: f32) i32
{
if (n <= 1) return 1;
var count: i32 = linearLogRecur(n / 2) +
linearLogRecur(n / 2);
var i: f32 = 0;
while (i < n) : (i += 1) {
count += 1;
}
return count;
}
```
![time_complexity_logarithmic_linear](time_complexity.assets/time_complexity_logarithmic_linear.png)
@ -2392,7 +2495,17 @@ $$
=== "Zig"
```zig title="time_complexity.zig"
// 阶乘阶（递归实现）
fn factorialRecur(n: i32) i32 {
if (n == 0) return 1;
var count: i32 = 0;
var i: i32 = 0;
// 从 1 个分裂出 n 个
while (i < n) : (i += 1) {
count += factorialRecur(n - 1);
}
return count;
}
```
![time_complexity_factorial](time_complexity.assets/time_complexity_factorial.png)
@ -2687,7 +2800,28 @@ $$
=== "Zig"
```zig title="worst_best_time_complexity.zig"
// 生成一个数组，元素为 { 1, 2, ..., n }，顺序被打乱
pub fn randomNumbers(comptime n: usize) [n]i32 {
var nums: [n]i32 = undefined;
// 生成数组 nums = { 1, 2, 3, ..., n }
for (nums) |*num, i| {
num.* = @intCast(i32, i) + 1;
}
// 随机打乱数组元素
const rand = std.crypto.random;
rand.shuffle(i32, &nums);
return nums;
}
// 查找数组 nums 中数字 1 所在索引
pub fn findOne(nums: []i32) i32 {
for (nums) |num, i| {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if (num == 1) return @intCast(i32, i);
}
return -1;
}
```
!!! tip