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feat: add solutions to lc problem: No.0239 #3894

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Dec 27, 2024
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feat: add solutions to lc problem: No.0239 #3894

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220 changes: 100 additions & 120 deletions solution/0200-0299/0239.Sliding Window Maximum/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -171,100 +171,6 @@ func (h *hp) Push(v any) { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
```

#### Rust

```rust
use std::collections::VecDeque;

impl Solution {
#[allow(dead_code)]
pub fn max_sliding_window(nums: Vec<i32>, k: i32) -> Vec<i32> {
// The deque contains the index of `nums`
let mut q: VecDeque<usize> = VecDeque::new();
let mut ans_vec: Vec<i32> = Vec::new();

for i in 0..nums.len() {
// Check the first element of queue, if it's out of bound
if !q.is_empty() && (i as i32) - k + 1 > (*q.front().unwrap() as i32) {
// Pop it out
q.pop_front();
}
// Pop back elements out until either the deque is empty
// Or the back element is greater than the current traversed element
while !q.is_empty() && nums[*q.back().unwrap()] <= nums[i] {
q.pop_back();
}
// Push the current index in queue
q.push_back(i);
// Check if the condition is satisfied
if i >= ((k - 1) as usize) {
ans_vec.push(nums[*q.front().unwrap()]);
}
}

ans_vec
}
}
```

#### JavaScript

```js
/**
* @param {number[]} nums
* @param {number} k
* @return {number[]}
*/
var maxSlidingWindow = function (nums, k) {
let ans = [];
let q = [];
for (let i = 0; i < nums.length; ++i) {
if (q && i - k + 1 > q[0]) {
q.shift();
}
while (q && nums[q[q.length - 1]] <= nums[i]) {
q.pop();
}
q.push(i);
if (i >= k - 1) {
ans.push(nums[q[0]]);
}
}
return ans;
};
```

#### C#

```cs
using System.Collections.Generic;

public class Solution {
public int[] MaxSlidingWindow(int[] nums, int k) {
if (nums.Length == 0) return new int[0];
var result = new int[nums.Length - k + 1];
var descOrderNums = new LinkedList<int>();
for (var i = 0; i < nums.Length; ++i)
{
if (i >= k && nums[i - k] == descOrderNums.First.Value)
{
descOrderNums.RemoveFirst();
}
while (descOrderNums.Count > 0 && nums[i] > descOrderNums.Last.Value)
{
descOrderNums.RemoveLast();
}
descOrderNums.AddLast(nums[i]);
if (i >= k - 1)
{
result[i - k + 1] = descOrderNums.First.Value;
}
}
return result;
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand All @@ -273,20 +179,11 @@ public class Solution {

### 方法二：单调队列

这道题也可以使用单调队列来解决。时间复杂度 $O(n)$，空间复杂度 $O(k)$。
求滑动窗口的最大值，一种常见的方法是使用单调队列。

单调队列常见模型：找出滑动窗口中的最大值/最小值。模板：
我们可以维护一个从队头到队尾单调递减的队列 $q$，队列中存储的是元素的下标。遍历数组 $\textit{nums}$，对于当前元素 $\textit{nums}[i]$，我们首先判断队头元素是否滑出窗口，如果滑出窗口则将队头元素弹出。然后我们将当前元素 $\textit{nums}[i]$ 从队尾开始依次与队尾元素比较，如果队尾元素小于等于当前元素，则将队尾元素弹出，直到队尾元素大于当前元素或者队列为空。然后将当前元素的下标加入队列。此时队列的队头元素即为当前滑动窗口的最大值，注意，我们将队头元素加入结果数组的时机是当下标 $i$ 大于等于 $k-1$ 时。

```python
q = deque()
for i in range(n):
# 判断队头是否滑出窗口
while q and checkout_out(q[0]):
q.popleft()
while q and check(q[-1]):
q.pop()
q.append(i)
```
时间复杂度 $O(n)$，空间复杂度 $O(k)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。

<!-- tabs:start -->

Expand All @@ -297,10 +194,10 @@ class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
q = deque()
ans = []
for i, v in enumerate(nums):
if q and i - k + 1 > q[0]:
for i, x in enumerate(nums):
if q and i - q[0] >= k:
q.popleft()
while q and nums[q[-1]] <= v:
while q and nums[q[-1]] <= x:
q.pop()
q.append(i)
if i >= k - 1:
Expand All @@ -316,16 +213,16 @@ class Solution {
int n = nums.length;
int[] ans = new int[n - k + 1];
Deque<Integer> q = new ArrayDeque<>();
for (int i = 0, j = 0; i < n; ++i) {
if (!q.isEmpty() && i - k + 1 > q.peekFirst()) {
for (int i = 0; i < n; ++i) {
if (!q.isEmpty() && i - q.peekFirst() >= k) {
q.pollFirst();
}
while (!q.isEmpty() && nums[q.peekLast()] <= nums[i]) {
q.pollLast();
}
q.offer(i);
q.offerLast(i);
if (i >= k - 1) {
ans[j++] = nums[q.peekFirst()];
ans[i - k + 1] = nums[q.peekFirst()];
}
}
return ans;
Expand All @@ -342,15 +239,15 @@ public:
deque<int> q;
vector<int> ans;
for (int i = 0; i < nums.size(); ++i) {
if (!q.empty() && i - k + 1 > q.front()) {
if (q.size() && i - q.front() >= k) {
q.pop_front();
}
while (!q.empty() && nums[q.back()] <= nums[i]) {
while (q.size() && nums[q.back()] <= nums[i]) {
q.pop_back();
}
q.push_back(i);
if (i >= k - 1) {
ans.emplace_back(nums[q.front()]);
ans.push_back(nums[q.front()]);
}
}
return ans;
Expand All @@ -363,22 +260,105 @@ public:
```go
func maxSlidingWindow(nums []int, k int) (ans []int) {
q := []int{}
for i, v := range nums {
if len(q) > 0 && i-k+1 > q[0] {
for i, x := range nums {
if len(q) > 0 && i-q[0] >= k {
q = q[1:]
}
for len(q) > 0 && nums[q[len(q)-1]] <= v {
for len(q) > 0 && nums[q[len(q)-1]] <= x {
q = q[:len(q)-1]
}
q = append(q, i)
if i >= k-1 {
ans = append(ans, nums[q[0]])
}
}
return ans
return
}
```

#### TypeScript

```ts
function maxSlidingWindow(nums: number[], k: number): number[] {
const ans: number[] = [];
const q = new Deque();
for (let i = 0; i < nums.length; ++i) {
if (!q.isEmpty() && i - q.front()! >= k) {
q.popFront();
}
while (!q.isEmpty() && nums[q.back()!] <= nums[i]) {
q.popBack();
}
q.pushBack(i);
if (i >= k - 1) {
ans.push(nums[q.front()!]);
}
}
return ans;
}
```

#### Rust

```rust
use std::collections::VecDeque;

impl Solution {
pub fn max_sliding_window(nums: Vec<i32>, k: i32) -> Vec<i32> {
let k = k as usize;
let mut ans = Vec::new();
let mut q: VecDeque<usize> = VecDeque::new();

for i in 0..nums.len() {
if let Some(&front) = q.front() {
if i >= front + k {
q.pop_front();
}
}
while let Some(&back) = q.back() {
if nums[back] <= nums[i] {
q.pop_back();
} else {
break;
}
}
q.push_back(i);
if i >= k - 1 {
ans.push(nums[*q.front().unwrap()]);
}
}
ans
}
}
```

#### JavaScript

```js
/**
* @param {number[]} nums
* @param {number} k
* @return {number[]}
*/
var maxSlidingWindow = function (nums, k) {
const ans = [];
const q = new Deque();
for (let i = 0; i < nums.length; ++i) {
if (!q.isEmpty() && i - q.front() >= k) {
q.popFront();
}
while (!q.isEmpty() && nums[q.back()] <= nums[i]) {
q.popBack();
}
q.pushBack(i);
if (i >= k - 1) {
ans.push(nums[q.front()]);
}
}
return ans;
};
```

<!-- tabs:end -->

<!-- solution:end -->
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