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Sep 13, 2024
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feat: update solutions to lc problem: No.2393 #3524

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163 changes: 23 additions & 140 deletions solution/2300-2399/2393.Count Strictly Increasing Subarrays/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -61,123 +61,15 @@ tags:

<!-- solution:start -->

### 方法一：双指针
### 方法一：枚举

利用双指针，找到每一段连续递增子数组的长度，我们记为 `cnt`，每次将 $(1+cnt)\times cnt / 2$ 累加到答案中。
我们可以枚举以每个元素结尾的严格递增子数组的个数，然后将它们累加起来即可。

时间复杂度 $O(n)$，空间复杂度 $O(1)$，其中 $n$ 是数组的长度。
我们用一个变量 $\textit{cnt}$ 来记录以当前元素结尾的严格递增子数组的个数，初始时 $\textit{cnt} = 1$。然后我们从第二个元素开始遍历数组，如果当前元素大于前一个元素，那么 $\textit{cnt}$ 就可以加 $1$，否则 $\textit{cnt}$ 重置为 $1$。此时，以当前元素结尾的严格递增子数组的个数就是 $\textit{cnt}$，我们将其累加到答案中即可。

<!-- tabs:start -->

#### Python3

```python
class Solution:
def countSubarrays(self, nums: List[int]) -> int:
ans = i = 0
while i < len(nums):
j = i + 1
while j < len(nums) and nums[j] > nums[j - 1]:
j += 1
cnt = j - i
ans += (1 + cnt) * cnt // 2
i = j
return ans
```

#### Java

```java
class Solution {
public long countSubarrays(int[] nums) {
long ans = 0;
int i = 0, n = nums.length;
while (i < n) {
int j = i + 1;
while (j < n && nums[j] > nums[j - 1]) {
++j;
}
long cnt = j - i;
ans += (1 + cnt) * cnt / 2;
i = j;
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
long long countSubarrays(vector<int>& nums) {
long long ans = 0;
int i = 0, n = nums.size();
while (i < n) {
int j = i + 1;
while (j < n && nums[j] > nums[j - 1]) {
++j;
}
int cnt = j - i;
ans += 1ll * (1 + cnt) * cnt / 2;
i = j;
}
return ans;
}
};
```
#### Go
```go
func countSubarrays(nums []int) int64 {
ans := 0
i, n := 0, len(nums)
for i < n {
j := i + 1
for j < n && nums[j] > nums[j-1] {
j++
}
cnt := j - i
ans += (1 + cnt) * cnt / 2
i = j
}
return int64(ans)
}
```

#### TypeScript

```ts
function countSubarrays(nums: number[]): number {
let ans = 0;
let i = 0;
const n = nums.length;
while (i < n) {
let j = i + 1;
while (j < n && nums[j] > nums[j - 1]) {
++j;
}
const cnt = j - i;
ans += ((1 + cnt) * cnt) / 2;
i = j;
}
return ans;
}
```

<!-- tabs:end -->
遍历结束后，返回答案即可。

<!-- solution:end -->

<!-- solution:start -->

### 方法二：枚举

我们可以枚举数组中的每一个元素，找到以该元素为结尾的严格递增子数组的个数，然后将这些个数累加到答案中。

时间复杂度 $O(n)$，空间复杂度 $O(1)$，其中 $n$ 是数组的长度。
时间复杂度 $O(n)$，其中 $n$ 为数组的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -186,13 +78,12 @@ function countSubarrays(nums: number[]): number {
```python
class Solution:
def countSubarrays(self, nums: List[int]) -> int:
ans = pre = cnt = 0
for x in nums:
if pre < x:
ans = cnt = 1
for x, y in pairwise(nums):
if x < y:
cnt += 1
else:
cnt = 1
pre = x
ans += cnt
return ans
```
Expand All @@ -202,15 +93,13 @@ class Solution:
```java
class Solution {
public long countSubarrays(int[] nums) {
long ans = 0;
int pre = 0, cnt = 0;
for (int x : nums) {
if (pre < x) {
long ans = 1, cnt = 1;
for (int i = 1; i < nums.length; ++i) {
if (nums[i - 1] < nums[i]) {
++cnt;
} else {
cnt = 1;
}
pre = x;
ans += cnt;
}
return ans;
Expand All @@ -224,16 +113,14 @@ class Solution {
class Solution {
public:
long long countSubarrays(vector<int>& nums) {
long long ans = 0;
int pre = 0, cnt = 0;
for (int x : nums) {
if (pre < x) {
long long ans = 1, cnt = 1;
for (int i = 1; i < nums.size(); ++i) {
if (nums[i - 1] < nums[i]) {
++cnt;
} else {
cnt = 1;
}
ans += cnt;
pre = x;
}
return ans;
}
Expand All @@ -243,36 +130,32 @@ public:
#### Go
```go
func countSubarrays(nums []int) (ans int64) {
pre, cnt := 0, 0
for _, x := range nums {
if pre < x {
func countSubarrays(nums []int) int64 {
ans, cnt := 1, 1
for i, x := range nums[1:] {
if nums[i] < x {
cnt++
} else {
cnt = 1
}
ans += int64(cnt)
pre = x
ans += cnt
}
return
return int64(ans)
}
```

#### TypeScript

```ts
function countSubarrays(nums: number[]): number {
let ans = 0;
let pre = 0;
let cnt = 0;
for (const x of nums) {
if (pre < x) {
let [ans, cnt] = [1, 1];
for (let i = 1; i < nums.length; ++i) {
if (nums[i - 1] < nums[i]) {
++cnt;
} else {
cnt = 1;
}
ans += cnt;
pre = x;
}
return ans;
}
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