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feat: add solutions to lc problem: No.3024 (#2310)
No.3024.Type of Triangle II
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‎solution/3000-3099/3024.Type of Triangle II/README.md

+79-4
Original file line numberDiff line numberDiff line change
@@ -50,24 +50,99 @@ nums[1] + nums[2] = 4 + 5 = 9 ，大于 nums[0] = 3 。
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## 解法
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### 方法一
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### 方法一：排序 + 分类讨论
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我们先对数组进行排序，然后根据三角形的定义进行分类讨论即可。
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- 如果最小的两个数之和小于等于最大的数，那么无法构成三角形，返回 "none"。
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- 如果最小的数等于最大的数，那么是等边三角形，返回 "equilateral"。
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- 如果最小的数等于中间的数或者中间的数等于最大的数，那么是等腰三角形，返回 "isosceles"。
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- 否则，返回 "scalene"。
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时间复杂度 $O(1)$，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 triangleType(self, nums: List[int]) -> str:
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nums.sort()
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if nums[0] + nums[1] <= nums[2]:
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return "none"
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if nums[0] == nums[2]:
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return "equilateral"
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if nums[0] == nums[1] or nums[1] == nums[2]:
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return "isosceles"
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return "scalene"
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```
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```java
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class Solution {
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public String triangleType(int[] nums) {
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Arrays.sort(nums);
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if (nums[0] + nums[1] <= nums[2]) {
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return "none";
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}
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if (nums[0] == nums[2]) {
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return "equilateral";
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}
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if (nums[0] == nums[1] || nums[1] == nums[2]) {
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return "isosceles";
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}
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return "scalene";
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}
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}
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```
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```cpp
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class Solution {
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public:
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string triangleType(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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if (nums[0] + nums[1] <= nums[2]) {
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return "none";
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}
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if (nums[0] == nums[2]) {
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return "equilateral";
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}
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if (nums[0] == nums[1] || nums[1] == nums[2]) {
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return "isosceles";
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}
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return "scalene";
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}
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};
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```
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```go
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func triangleType(nums []int) string {
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sort.Ints(nums)
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if nums[0]+nums[1] <= nums[2] {
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return "none"
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}
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if nums[0] == nums[2] {
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return "equilateral"
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}
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if nums[0] == nums[1] || nums[1] == nums[2] {
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return "isosceles"
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}
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return "scalene"
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}
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```
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```ts
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function triangleType(nums: number[]): string {
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nums.sort((a, b) => a - b);
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if (nums[0] + nums[1] <= nums[2]) {
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return 'none';
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}
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if (nums[0] === nums[2]) {
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return 'equilateral';
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}
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if (nums[0] === nums[1] || nums[1] === nums[2]) {
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return 'isosceles';
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}
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return 'scalene';
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}
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```
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<!-- tabs:end -->

‎solution/3000-3099/3024.Type of Triangle II/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -46,24 +46,99 @@ As all the sides are of different lengths, it will form a scalene triangle.
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## Solutions
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### Solution 1
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### Solution 1: Sorting + Case Discussion
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First, we sort the array, and then we can classify and discuss according to the definition of a triangle.
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- If the sum of the smallest two numbers is less than or equal to the largest number, then it cannot form a triangle, return "none".
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- If the smallest number is equal to the largest number, then it is an equilateral triangle, return "equilateral".
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- If the smallest number is equal to the middle number or the middle number is equal to the largest number, then it is an isosceles triangle, return "isosceles".
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- Otherwise, return "scalene".
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The time complexity is $O(1)$, and the space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def triangleType(self, nums: List[int]) -> str:
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nums.sort()
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if nums[0] + nums[1] <= nums[2]:
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return "none"
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if nums[0] == nums[2]:
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return "equilateral"
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if nums[0] == nums[1] or nums[1] == nums[2]:
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return "isosceles"
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return "scalene"
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```
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```java
58-
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class Solution {
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public String triangleType(int[] nums) {
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Arrays.sort(nums);
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if (nums[0] + nums[1] <= nums[2]) {
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return "none";
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}
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if (nums[0] == nums[2]) {
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return "equilateral";
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}
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if (nums[0] == nums[1] || nums[1] == nums[2]) {
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return "isosceles";
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}
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return "scalene";
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}
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}
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```
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```cpp
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class Solution {
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public:
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string triangleType(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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if (nums[0] + nums[1] <= nums[2]) {
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return "none";
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}
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if (nums[0] == nums[2]) {
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return "equilateral";
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}
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if (nums[0] == nums[1] || nums[1] == nums[2]) {
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return "isosceles";
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}
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return "scalene";
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}
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};
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```
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```go
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func triangleType(nums []int) string {
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sort.Ints(nums)
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if nums[0]+nums[1] <= nums[2] {
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return "none"
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}
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if nums[0] == nums[2] {
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return "equilateral"
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}
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if nums[0] == nums[1] || nums[1] == nums[2] {
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return "isosceles"
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}
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return "scalene"
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}
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```
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128+
```ts
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function triangleType(nums: number[]): string {
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nums.sort((a, b) => a - b);
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if (nums[0] + nums[1] <= nums[2]) {
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return 'none';
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}
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if (nums[0] === nums[2]) {
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return 'equilateral';
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}
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if (nums[0] === nums[1] || nums[1] === nums[2]) {
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return 'isosceles';
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}
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return 'scalene';
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}
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```
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<!-- tabs:end -->

‎solution/3000-3099/3024.Type of Triangle II/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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class Solution {
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public:
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string triangleType(vector<int>& nums) {
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sort(nums.begin(), nums.end());
5+
if (nums[0] + nums[1] <= nums[2]) {
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return "none";
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}
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if (nums[0] == nums[2]) {
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return "equilateral";
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}
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if (nums[0] == nums[1] || nums[1] == nums[2]) {
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return "isosceles";
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}
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return "scalene";
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}
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};

‎solution/3000-3099/3024.Type of Triangle II/Solution.go

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func triangleType(nums []int) string {
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sort.Ints(nums)
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if nums[0]+nums[1] <= nums[2] {
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return "none"
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}
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if nums[0] == nums[2] {
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return "equilateral"
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}
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if nums[0] == nums[1] || nums[1] == nums[2] {
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return "isosceles"
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}
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return "scalene"
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}

‎solution/3000-3099/3024.Type of Triangle II/Solution.java

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class Solution {
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public String triangleType(int[] nums) {
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Arrays.sort(nums);
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if (nums[0] + nums[1] <= nums[2]) {
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return "none";
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}
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if (nums[0] == nums[2]) {
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return "equilateral";
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}
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if (nums[0] == nums[1] || nums[1] == nums[2]) {
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return "isosceles";
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}
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return "scalene";
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}
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}

‎solution/3000-3099/3024.Type of Triangle II/Solution.py

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class Solution:
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def triangleType(self, nums: List[int]) -> str:
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nums.sort()
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if nums[0] + nums[1] <= nums[2]:
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return "none"
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if nums[0] == nums[2]:
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return "equilateral"
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if nums[0] == nums[1] or nums[1] == nums[2]:
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return "isosceles"
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return "scalene"

‎solution/3000-3099/3024.Type of Triangle II/Solution.ts

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function triangleType(nums: number[]): string {
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nums.sort((a, b) => a - b);
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if (nums[0] + nums[1] <= nums[2]) {
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return 'none';
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}
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if (nums[0] === nums[2]) {
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return 'equilateral';
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}
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if (nums[0] === nums[1] || nums[1] === nums[2]) {
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return 'isosceles';
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}
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return 'scalene';
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}

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