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Commit fb7ac1c

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feat: add solutions to lc problem: No.0239 (doocs#3894)
No.0239.Sliding Window Maximum
1 parent a471399 commit fb7ac1c

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-327
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‎solution/0200-0299/0239.Sliding Window Maximum/README.md‎

Lines changed: 100 additions & 120 deletions
Original file line numberDiff line numberDiff line change
@@ -171,100 +171,6 @@ func (h *hp) Push(v any) { *h = append(*h, v.(pair)) }
171171
func (h *hp) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
172172
```
173173

174-
#### Rust
175-
176-
```rust
177-
use std::collections::VecDeque;
178-
179-
impl Solution {
180-
#[allow(dead_code)]
181-
pub fn max_sliding_window(nums: Vec<i32>, k: i32) -> Vec<i32> {
182-
// The deque contains the index of `nums`
183-
let mut q: VecDeque<usize> = VecDeque::new();
184-
let mut ans_vec: Vec<i32> = Vec::new();
185-
186-
for i in 0..nums.len() {
187-
// Check the first element of queue, if it's out of bound
188-
if !q.is_empty() && (i as i32) - k + 1 > (*q.front().unwrap() as i32) {
189-
// Pop it out
190-
q.pop_front();
191-
}
192-
// Pop back elements out until either the deque is empty
193-
// Or the back element is greater than the current traversed element
194-
while !q.is_empty() && nums[*q.back().unwrap()] <= nums[i] {
195-
q.pop_back();
196-
}
197-
// Push the current index in queue
198-
q.push_back(i);
199-
// Check if the condition is satisfied
200-
if i >= ((k - 1) as usize) {
201-
ans_vec.push(nums[*q.front().unwrap()]);
202-
}
203-
}
204-
205-
ans_vec
206-
}
207-
}
208-
```
209-
210-
#### JavaScript
211-
212-
```js
213-
/**
214-
* @param {number[]} nums
215-
* @param {number} k
216-
* @return {number[]}
217-
*/
218-
var maxSlidingWindow = function (nums, k) {
219-
let ans = [];
220-
let q = [];
221-
for (let i = 0; i < nums.length; ++i) {
222-
if (q && i - k + 1 > q[0]) {
223-
q.shift();
224-
}
225-
while (q && nums[q[q.length - 1]] <= nums[i]) {
226-
q.pop();
227-
}
228-
q.push(i);
229-
if (i >= k - 1) {
230-
ans.push(nums[q[0]]);
231-
}
232-
}
233-
return ans;
234-
};
235-
```
236-
237-
#### C#
238-
239-
```cs
240-
using System.Collections.Generic;
241-
242-
public class Solution {
243-
public int[] MaxSlidingWindow(int[] nums, int k) {
244-
if (nums.Length == 0) return new int[0];
245-
var result = new int[nums.Length - k + 1];
246-
var descOrderNums = new LinkedList<int>();
247-
for (var i = 0; i < nums.Length; ++i)
248-
{
249-
if (i >= k && nums[i - k] == descOrderNums.First.Value)
250-
{
251-
descOrderNums.RemoveFirst();
252-
}
253-
while (descOrderNums.Count > 0 && nums[i] > descOrderNums.Last.Value)
254-
{
255-
descOrderNums.RemoveLast();
256-
}
257-
descOrderNums.AddLast(nums[i]);
258-
if (i >= k - 1)
259-
{
260-
result[i - k + 1] = descOrderNums.First.Value;
261-
}
262-
}
263-
return result;
264-
}
265-
}
266-
```
267-
268174
<!-- tabs:end -->
269175

270176
<!-- solution:end -->
@@ -273,20 +179,11 @@ public class Solution {
273179

274180
### 方法二:单调队列
275181

276-
这道题也可以使用单调队列来解决。时间复杂度 $O(n),ドル空间复杂度 $O(k)$。
182+
求滑动窗口的最大值,一种常见的方法是使用单调队列。
277183

278-
单调队列常见模型:找出滑动窗口中的最大值/最小值。模板:
184+
我们可以维护一个从队头到队尾单调递减的队列 $q,ドル队列中存储的是元素的下标。遍历数组 $\textit{nums},ドル对于当前元素 $\textit{nums}[i],ドル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。
279185

280-
```python
281-
q = deque()
282-
for i in range(n):
283-
# 判断队头是否滑出窗口
284-
while q and checkout_out(q[0]):
285-
q.popleft()
286-
while q and check(q[-1]):
287-
q.pop()
288-
q.append(i)
289-
```
186+
时间复杂度 $O(n),ドル空间复杂度 $O(k)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
290187

291188
<!-- tabs:start -->
292189

@@ -297,10 +194,10 @@ class Solution:
297194
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
298195
q = deque()
299196
ans = []
300-
for i, v in enumerate(nums):
301-
if q and i - k +1>q[0]:
197+
for i, x in enumerate(nums):
198+
if q and i - q[0]>= k:
302199
q.popleft()
303-
while q and nums[q[-1]] <= v:
200+
while q and nums[q[-1]] <= x:
304201
q.pop()
305202
q.append(i)
306203
if i >= k - 1:
@@ -316,16 +213,16 @@ class Solution {
316213
int n = nums.length;
317214
int[] ans = new int[n - k + 1];
318215
Deque<Integer> q = new ArrayDeque<>();
319-
for (int i = 0, j =0; i < n; ++i) {
320-
if (!q.isEmpty() && i - k +1>q.peekFirst()) {
216+
for (int i = 0; i < n; ++i) {
217+
if (!q.isEmpty() && i - q.peekFirst()>= k) {
321218
q.pollFirst();
322219
}
323220
while (!q.isEmpty() && nums[q.peekLast()] <= nums[i]) {
324221
q.pollLast();
325222
}
326-
q.offer(i);
223+
q.offerLast(i);
327224
if (i >= k - 1) {
328-
ans[j++] = nums[q.peekFirst()];
225+
ans[i - k +1] = nums[q.peekFirst()];
329226
}
330227
}
331228
return ans;
@@ -342,15 +239,15 @@ public:
342239
deque<int> q;
343240
vector<int> ans;
344241
for (int i = 0; i < nums.size(); ++i) {
345-
if (!q.empty() && i - k + 1 > q.front()) {
242+
if (q.size() && i - q.front() >= k) {
346243
q.pop_front();
347244
}
348-
while (!q.empty() && nums[q.back()] <= nums[i]) {
245+
while (q.size() && nums[q.back()] <= nums[i]) {
349246
q.pop_back();
350247
}
351248
q.push_back(i);
352249
if (i >= k - 1) {
353-
ans.emplace_back(nums[q.front()]);
250+
ans.push_back(nums[q.front()]);
354251
}
355252
}
356253
return ans;
@@ -363,22 +260,105 @@ public:
363260
```go
364261
func maxSlidingWindow(nums []int, k int) (ans []int) {
365262
q := []int{}
366-
for i, v := range nums {
367-
if len(q) > 0 && i-k+1 > q[0] {
263+
for i, x := range nums {
264+
if len(q) > 0 && i-q[0] >= k {
368265
q = q[1:]
369266
}
370-
for len(q) > 0 && nums[q[len(q)-1]] <= v {
267+
for len(q) > 0 && nums[q[len(q)-1]] <= x {
371268
q = q[:len(q)-1]
372269
}
373270
q = append(q, i)
374271
if i >= k-1 {
375272
ans = append(ans, nums[q[0]])
376273
}
377274
}
378-
return ans
275+
return
276+
}
277+
```
278+
279+
#### TypeScript
280+
281+
```ts
282+
function maxSlidingWindow(nums: number[], k: number): number[] {
283+
const ans: number[] = [];
284+
const q = new Deque();
285+
for (let i = 0; i < nums.length; ++i) {
286+
if (!q.isEmpty() && i - q.front()! >= k) {
287+
q.popFront();
288+
}
289+
while (!q.isEmpty() && nums[q.back()!] <= nums[i]) {
290+
q.popBack();
291+
}
292+
q.pushBack(i);
293+
if (i >= k - 1) {
294+
ans.push(nums[q.front()!]);
295+
}
296+
}
297+
return ans;
298+
}
299+
```
300+
301+
#### Rust
302+
303+
```rust
304+
use std::collections::VecDeque;
305+
306+
impl Solution {
307+
pub fn max_sliding_window(nums: Vec<i32>, k: i32) -> Vec<i32> {
308+
let k = k as usize;
309+
let mut ans = Vec::new();
310+
let mut q: VecDeque<usize> = VecDeque::new();
311+
312+
for i in 0..nums.len() {
313+
if let Some(&front) = q.front() {
314+
if i >= front + k {
315+
q.pop_front();
316+
}
317+
}
318+
while let Some(&back) = q.back() {
319+
if nums[back] <= nums[i] {
320+
q.pop_back();
321+
} else {
322+
break;
323+
}
324+
}
325+
q.push_back(i);
326+
if i >= k - 1 {
327+
ans.push(nums[*q.front().unwrap()]);
328+
}
329+
}
330+
ans
331+
}
379332
}
380333
```
381334

335+
#### JavaScript
336+
337+
```js
338+
/**
339+
* @param {number[]} nums
340+
* @param {number} k
341+
* @return {number[]}
342+
*/
343+
var maxSlidingWindow = function (nums, k) {
344+
const ans = [];
345+
const q = new Deque();
346+
for (let i = 0; i < nums.length; ++i) {
347+
if (!q.isEmpty() && i - q.front() >= k) {
348+
q.popFront();
349+
}
350+
while (!q.isEmpty() && nums[q.back()] <= nums[i]) {
351+
q.popBack();
352+
}
353+
q.pushBack(i);
354+
if (i >= k - 1) {
355+
ans.push(nums[q.front()]);
356+
}
357+
}
358+
return ans;
359+
};
360+
```
361+
382362
<!-- tabs:end -->
383363

384364
<!-- solution:end -->

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