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Commit 92eac10

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feat: add solutions to lc problem: No.3637 (doocs#4618)
No.3637.Trionic Array I
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‎solution/3600-3699/3637.Trionic Array I/README.md‎

Lines changed: 118 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -73,32 +73,146 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3637.Tr
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<!-- solution:start -->
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### 方法一
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### 方法一:一次遍历
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我们首先定义一个指针 $p,ドル初始时 $p = 0,ドル表示当前指向数组的第一个元素。我们将 $p$ 向右移动,直到找到第一个不满足严格递增的元素,即 $nums[p] \geq nums[p + 1]$。如果此时 $p = 0,ドル说明数组的前半部分没有严格递增的部分,因此直接返回 $\text{false}$。
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接下来,我们定义另一个指针 $q,ドル初始时 $q = p,ドル表示当前指向数组的第二个部分的第一个元素。我们将 $q$ 向右移动,直到找到第一个不满足严格递减的元素,即 $nums[q] \leq nums[q + 1]$。如果此时 $q = p$ 或者 $q = n - 1,ドル说明数组的第二部分没有严格递减的部分或者没有第三部分,因此直接返回 $\text{false}$。
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如果以上条件都满足,说明数组是三段式的,返回 $\text{true}$。
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时间复杂度 $O(n),ドル其中 $n$ 是数组的长度。空间复杂度 $O(1),ドル只使用了常数级别的额外空间。
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<!-- tabs:start -->
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#### Python3
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```python
83-
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class Solution:
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def isTrionic(self, nums: List[int]) -> bool:
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n = len(nums)
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p = 0
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while p < n - 2 and nums[p] < nums[p + 1]:
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p += 1
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if p == 0:
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return False
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q = p
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while q < n - 1 and nums[q] > nums[q + 1]:
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q += 1
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if q == p or q == n - 1:
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return False
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while q < n - 1 and nums[q] < nums[q + 1]:
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q += 1
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return q == n - 1
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```
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#### Java
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```java
89-
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class Solution {
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public boolean isTrionic(int[] nums) {
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int n = nums.length;
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int p = 0;
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while (p < n - 2 && nums[p] < nums[p + 1]) {
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p++;
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}
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if (p == 0) {
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return false;
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}
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int q = p;
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while (q < n - 1 && nums[q] > nums[q + 1]) {
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q++;
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}
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if (q == p || q == n - 1) {
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return false;
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}
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while (q < n - 1 && nums[q] < nums[q + 1]) {
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q++;
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}
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return q == n - 1;
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}
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}
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```
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#### C++
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```cpp
95-
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class Solution {
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public:
142+
bool isTrionic(vector<int>& nums) {
143+
int n = nums.size();
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int p = 0;
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while (p < n - 2 && nums[p] < nums[p + 1]) {
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p++;
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}
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if (p == 0) {
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return false;
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}
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int q = p;
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while (q < n - 1 && nums[q] > nums[q + 1]) {
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q++;
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}
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if (q == p || q == n - 1) {
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return false;
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}
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while (q < n - 1 && nums[q] < nums[q + 1]) {
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q++;
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}
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return q == n - 1;
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}
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};
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```
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#### Go
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```go
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func isTrionic(nums []int) bool {
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n := len(nums)
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p := 0
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for p < n-2 && nums[p] < nums[p+1] {
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p++
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}
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if p == 0 {
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return false
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}
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q := p
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for q < n-1 && nums[q] > nums[q+1] {
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q++
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}
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if q == p || q == n-1 {
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return false
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}
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for q < n-1 && nums[q] < nums[q+1] {
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q++
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}
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return q == n-1
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}
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```
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#### TypeScript
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```ts
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function isTrionic(nums: number[]): boolean {
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const n = nums.length;
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let p = 0;
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while (p < n - 2 && nums[p] < nums[p + 1]) {
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p++;
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}
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if (p === 0) {
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return false;
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}
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let q = p;
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while (q < n - 1 && nums[q] > nums[q + 1]) {
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q++;
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}
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if (q === p || q === n - 1) {
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return false;
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}
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while (q < n - 1 && nums[q] < nums[q + 1]) {
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q++;
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}
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return q === n - 1;
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}
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```
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<!-- tabs:end -->

‎solution/3600-3699/3637.Trionic Array I/README_EN.md‎

Lines changed: 118 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -71,32 +71,146 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3637.Tr
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Single Pass
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We first define a pointer $p,ドル initially $p = 0,ドル pointing to the first element of the array. We move $p$ to the right until we find the first element that doesn't satisfy strict increasing order, i.e., $nums[p] \geq nums[p + 1]$. If $p = 0$ at this point, it means the first part of the array doesn't have a strictly increasing section, so we return $\text{false}$ directly.
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Next, we define another pointer $q,ドル initially $q = p,ドル pointing to the first element of the second part of the array. We move $q$ to the right until we find the first element that doesn't satisfy strict decreasing order, i.e., $nums[q] \leq nums[q + 1]$. If $q = p$ or $q = n - 1$ at this point, it means the second part of the array doesn't have a strictly decreasing section or there's no third part, so we return $\text{false}$ directly.
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If all the above conditions are satisfied, it means the array is trionic, and we return $\text{true}$.
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The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(1),ドル using only constant extra space.
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<!-- tabs:start -->
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#### Python3
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```python
81-
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class Solution:
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def isTrionic(self, nums: List[int]) -> bool:
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n = len(nums)
92+
p = 0
93+
while p < n - 2 and nums[p] < nums[p + 1]:
94+
p += 1
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if p == 0:
96+
return False
97+
q = p
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while q < n - 1 and nums[q] > nums[q + 1]:
99+
q += 1
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if q == p or q == n - 1:
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return False
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while q < n - 1 and nums[q] < nums[q + 1]:
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q += 1
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return q == n - 1
82105
```
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#### Java
85108

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```java
87-
110+
class Solution {
111+
public boolean isTrionic(int[] nums) {
112+
int n = nums.length;
113+
int p = 0;
114+
while (p < n - 2 && nums[p] < nums[p + 1]) {
115+
p++;
116+
}
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if (p == 0) {
118+
return false;
119+
}
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int q = p;
121+
while (q < n - 1 && nums[q] > nums[q + 1]) {
122+
q++;
123+
}
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if (q == p || q == n - 1) {
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return false;
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}
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while (q < n - 1 && nums[q] < nums[q + 1]) {
128+
q++;
129+
}
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return q == n - 1;
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}
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}
88133
```
89134

90135
#### C++
91136

92137
```cpp
93-
138+
class Solution {
139+
public:
140+
bool isTrionic(vector<int>& nums) {
141+
int n = nums.size();
142+
int p = 0;
143+
while (p < n - 2 && nums[p] < nums[p + 1]) {
144+
p++;
145+
}
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if (p == 0) {
147+
return false;
148+
}
149+
int q = p;
150+
while (q < n - 1 && nums[q] > nums[q + 1]) {
151+
q++;
152+
}
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if (q == p || q == n - 1) {
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return false;
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}
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while (q < n - 1 && nums[q] < nums[q + 1]) {
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q++;
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}
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return q == n - 1;
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}
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};
94162
```
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#### Go
97165
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```go
167+
func isTrionic(nums []int) bool {
168+
n := len(nums)
169+
p := 0
170+
for p < n-2 && nums[p] < nums[p+1] {
171+
p++
172+
}
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if p == 0 {
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return false
175+
}
176+
q := p
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for q < n-1 && nums[q] > nums[q+1] {
178+
q++
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}
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if q == p || q == n-1 {
181+
return false
182+
}
183+
for q < n-1 && nums[q] < nums[q+1] {
184+
q++
185+
}
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return q == n-1
187+
}
188+
```
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190+
#### TypeScript
191+
192+
```ts
193+
function isTrionic(nums: number[]): boolean {
194+
const n = nums.length;
195+
let p = 0;
196+
while (p < n - 2 && nums[p] < nums[p + 1]) {
197+
p++;
198+
}
199+
if (p === 0) {
200+
return false;
201+
}
202+
let q = p;
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while (q < n - 1 && nums[q] > nums[q + 1]) {
204+
q++;
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}
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if (q === p || q === n - 1) {
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return false;
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}
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while (q < n - 1 && nums[q] < nums[q + 1]) {
210+
q++;
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}
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return q === n - 1;
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}
100214
```
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<!-- tabs:end -->
Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
1+
class Solution {
2+
public:
3+
bool isTrionic(vector<int>& nums) {
4+
int n = nums.size();
5+
int p = 0;
6+
while (p < n - 2 && nums[p] < nums[p + 1]) {
7+
p++;
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}
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if (p == 0) {
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return false;
11+
}
12+
int q = p;
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while (q < n - 1 && nums[q] > nums[q + 1]) {
14+
q++;
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}
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if (q == p || q == n - 1) {
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return false;
18+
}
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while (q < n - 1 && nums[q] < nums[q + 1]) {
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q++;
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}
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return q == n - 1;
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}
24+
};
Lines changed: 21 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,21 @@
1+
func isTrionic(nums []int) bool {
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n := len(nums)
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p := 0
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for p < n-2 && nums[p] < nums[p+1] {
5+
p++
6+
}
7+
if p == 0 {
8+
return false
9+
}
10+
q := p
11+
for q < n-1 && nums[q] > nums[q+1] {
12+
q++
13+
}
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if q == p || q == n-1 {
15+
return false
16+
}
17+
for q < n-1 && nums[q] < nums[q+1] {
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q++
19+
}
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return q == n-1
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}
Lines changed: 23 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,23 @@
1+
class Solution {
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public boolean isTrionic(int[] nums) {
3+
int n = nums.length;
4+
int p = 0;
5+
while (p < n - 2 && nums[p] < nums[p + 1]) {
6+
p++;
7+
}
8+
if (p == 0) {
9+
return false;
10+
}
11+
int q = p;
12+
while (q < n - 1 && nums[q] > nums[q + 1]) {
13+
q++;
14+
}
15+
if (q == p || q == n - 1) {
16+
return false;
17+
}
18+
while (q < n - 1 && nums[q] < nums[q + 1]) {
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q++;
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}
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return q == n - 1;
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}
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}
Lines changed: 16 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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class Solution:
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def isTrionic(self, nums: List[int]) -> bool:
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n = len(nums)
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p = 0
5+
while p < n - 2 and nums[p] < nums[p + 1]:
6+
p += 1
7+
if p == 0:
8+
return False
9+
q = p
10+
while q < n - 1 and nums[q] > nums[q + 1]:
11+
q += 1
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if q == p or q == n - 1:
13+
return False
14+
while q < n - 1 and nums[q] < nums[q + 1]:
15+
q += 1
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return q == n - 1

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