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feat: add solutions to lc problem: No.3034 (doocs#2369)
No.3034.Number of Subarrays That Match a Pattern I
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‎solution/3000-3099/3034.Number of Subarrays That Match a Pattern I/README.md

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@@ -53,52 +53,47 @@
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## 解法
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### 方法一
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### 方法一：枚举
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我们可以枚举数组 `nums` 的所有长度为 $m + 1$ 的子数组，然后判断是否满足模式数组 `pattern`，如果满足则答案加一。
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时间复杂度 $O(n \times m)$，其中 $n$ 和 $m$ 分别是数组 `nums` 和 `pattern` 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def countMatchingSubarrays(self, nums: List[int], pattern: List[int]) -> int:
63-
n = len(nums)
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m = len(pattern)
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count = 0
66-
for i in range(n - m):
67-
flag = True
68-
for j in range(m):
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if (
70-
(pattern[j] == 1 and nums[i + j + 1] <= nums[i + j])
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or (pattern[j] == 0 and nums[i + j + 1] != nums[i + j])
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or (pattern[j] == -1 and nums[i + j + 1] >= nums[i + j])
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):
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flag = False
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break
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if flag:
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count += 1
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return count
67+
def f(a: int, b: int) -> int:
68+
return 0 if a == b else (1 if a < b else -1)
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ans = 0
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for i in range(len(nums) - len(pattern)):
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ans += all(
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f(nums[i + k], nums[i + k + 1]) == p for k, p in enumerate(pattern)
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)
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return ans
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```
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```java
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class Solution {
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public int countMatchingSubarrays(int[] nums, int[] pattern) {
84-
int n = nums.length;
85-
int m = pattern.length;
86-
int count = 0;
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for (int i = 0; i <= n - m - 1; i++) {
88-
boolean flag = true;
89-
for (int j = 0; j < m; j++) {
90-
if ((pattern[j] == 1 && nums[i + j + 1] <= nums[i + j]) ||
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(pattern[j] == 0 && nums[i + j + 1] != nums[i + j]) ||
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(pattern[j] == -1 && nums[i + j + 1] >= nums[i + j])) {
93-
flag = false;
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break;
81+
int n = nums.length, m = pattern.length;
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int ans = 0;
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for (int i = 0; i < n - m; ++i) {
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int ok = 1;
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for (int k = 0; k < m && ok == 1; ++k) {
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if (f(nums[i + k], nums[i + k + 1]) != pattern[k]) {
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ok = 0;
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}
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}
97-
if (flag) {
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count++;
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}
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ans += ok;
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}
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return count;
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return ans;
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}
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private int f(int a, int b) {
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return a == b ? 0 : (a < b ? 1 : -1);
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}
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}
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```
@@ -107,71 +102,89 @@ class Solution {
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class Solution {
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public:
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int countMatchingSubarrays(vector<int>& nums, vector<int>& pattern) {
110-
int n = nums.size();
111-
int m = pattern.size();
112-
int c = 0;
113-
for (int i = 0; i <= n - m - 1; i++) {
114-
bool flag = true;
115-
for (int j = 0; j < m; j++) {
116-
if ((pattern[j] == 1 && nums[i + j + 1] <= nums[i + j]) || (pattern[j] == 0 && nums[i + j + 1] != nums[i + j]) || (pattern[j] == -1 && nums[i + j + 1] >= nums[i + j])) {
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flag = false;
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break;
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int n = nums.size(), m = pattern.size();
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int ans = 0;
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auto f = [](int a, int b) {
108+
return a == b ? 0 : (a < b ? 1 : -1);
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};
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for (int i = 0; i < n - m; ++i) {
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int ok = 1;
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for (int k = 0; k < m && ok == 1; ++k) {
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if (f(nums[i + k], nums[i + k + 1]) != pattern[k]) {
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ok = 0;
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}
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}
121-
if (flag) {
122-
c++;
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}
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ans += ok;
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}
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return c;
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return ans;
126120
}
127121
};
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```
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```go
131-
func countMatchingSubarrays(nums []int, pattern []int) int {
132-
n := len(nums)
133-
m := len(pattern)
134-
count := 0
135-
for i := 0; i <= n-m-1; i++ {
136-
flag := true
137-
for j := 0; j < m; j++ {
138-
if (pattern[j] == 1 && nums[i+j+1] <= nums[i+j]) ||
139-
(pattern[j] == 0 && nums[i+j+1] != nums[i+j]) ||
140-
(pattern[j] == -1 && nums[i+j+1] >= nums[i+j]) {
141-
flag = false
142-
break
143-
}
125+
func countMatchingSubarrays(nums []int, pattern []int) (ans int) {
126+
f := func(a, b int) int {
127+
if a == b {
128+
return 0
144129
}
145-
if flag {
146-
count++
130+
if a < b {
131+
return 1
132+
}
133+
return -1
134+
}
135+
n, m := len(nums), len(pattern)
136+
for i := 0; i < n-m; i++ {
137+
ok := 1
138+
for k := 0; k < m && ok == 1; k++ {
139+
if f(nums[i+k], nums[i+k+1]) != pattern[k] {
140+
ok = 0
141+
}
147142
}
143+
ans += ok
148144
}
149-
return count
145+
return
150146
}
151147
```
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```ts
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function countMatchingSubarrays(nums: number[], pattern: number[]): number {
155-
const n: number = nums.length;
156-
const m: number = pattern.length;
157-
let count: number = 0;
158-
for (let i = 0; i <= n - m - 1; i++) {
159-
let flag: boolean = true;
160-
for (let j = 0; j < m; j++) {
161-
if (
162-
(pattern[j] === 1 && nums[i + j + 1] <= nums[i + j]) ||
163-
(pattern[j] === 0 && nums[i + j + 1] !== nums[i + j]) ||
164-
(pattern[j] === -1 && nums[i + j + 1] >= nums[i + j])
165-
) {
166-
flag = false;
167-
break;
151+
const f = (a: number, b: number) => (a === b ? 0 : a < b ? 1 : -1);
152+
const n = nums.length;
153+
const m = pattern.length;
154+
let ans = 0;
155+
for (let i = 0; i < n - m; ++i) {
156+
let ok = 1;
157+
for (let k = 0; k < m && ok; ++k) {
158+
if (f(nums[i + k], nums[i + k + 1]) !== pattern[k]) {
159+
ok = 0;
168160
}
169161
}
170-
if (flag) {
171-
count++;
162+
ans += ok;
163+
}
164+
return ans;
165+
}
166+
```
167+
168+
```cs
169+
public class Solution {
170+
public int CountMatchingSubarrays(int[] nums, int[] pattern) {
171+
int n = nums.Length, m = pattern.Length;
172+
int ans = 0;
173+
for (int i = 0; i < n - m; ++i) {
174+
int ok = 1;
175+
for (int k = 0; k < m && ok == 1; ++k) {
176+
if (f(nums[i + k], nums[i + k + 1]) != pattern[k]) {
177+
ok = 0;
178+
}
179+
}
180+
ans += ok;
172181
}
182+
return ans;
183+
}
184+
185+
private int f(int a, int b) {
186+
return a == b ? 0 : (a < b ? 1 : -1);
173187
}
174-
return count;
175188
}
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```
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