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Jun 28, 2024
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feat: add solutions to lc problems: No.0349,2285 #3171

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77 changes: 21 additions & 56 deletions solution/0300-0399/0349.Intersection of Two Arrays/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -134,6 +134,15 @@ func intersection(nums1 []int, nums2 []int) (ans []int) {
}
```

#### TypeScript

```ts
function intersection(nums1: number[], nums2: number[]): number[] {
const s = new Set(nums1);
return [...new Set(nums2.filter(x => s.has(x)))];
}
```

#### JavaScript

```js
Expand All @@ -143,18 +152,8 @@ func intersection(nums1 []int, nums2 []int) (ans []int) {
* @return {number[]}
*/
var intersection = function (nums1, nums2) {
const s = Array(1001).fill(false);
for (const x of nums1) {
s[x] = true;
}
const ans = [];
for (const x of nums2) {
if (s[x]) {
ans.push(x);
s[x] = false;
}
}
return ans;
const s = new Set(nums1);
return [...new Set(nums2.filter(x => s.has(x)))];
};
```

Expand All @@ -163,15 +162,12 @@ var intersection = function (nums1, nums2) {
```cs
public class Solution {
public int[] Intersection(int[] nums1, int[] nums2) {
List<int> result = new List<int>();
HashSet<int> arr1 = new(nums1);
HashSet<int> arr2 = new(nums2);
foreach (int x in arr1) {
if (arr2.Contains(x)) {
result.Add(x);
}
}
return result.ToArray();
HashSet<int> s1 = new HashSet<int>(nums1);
HashSet<int> s2 = new HashSet<int>(nums2);
s1.IntersectWith(s2);
int[] ans = new int[s1.Count];
s1.CopyTo(ans);
return ans;
}
}
```
Expand All @@ -186,18 +182,10 @@ class Solution {
* @return Integer[]
*/
function intersection($nums1, $nums2) {
$rs = [];
$set1 = array_values(array_unique($nums1));
$set2 = array_values(array_unique($nums2));
for ($i = 0; $i < count($set1); $i++) {
$hashmap[$set1[$i]] = 1;
}
for ($j = 0; $j < count($set2); $j++) {
if ($hashmap[$set2[$j]]) {
array_push($rs, $set2[$j]);
}
}
return $rs;
$s1 = array_unique($nums1);
$s2 = array_unique($nums2);
$ans = array_intersect($s1, $s2);
return array_values($ans);
}
}
```
Expand All @@ -206,27 +194,4 @@ class Solution {

<!-- solution:end -->

<!-- solution:start -->

### 方法二

<!-- tabs:start -->

#### JavaScript

```js
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number[]}
*/
var intersection = function (nums1, nums2) {
return Array.from(new Set(nums1)).filter(num => new Set(nums2).has(num));
};
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
85 changes: 28 additions & 57 deletions solution/0300-0399/0349.Intersection of Two Arrays/README_EN.md
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Expand Up @@ -52,7 +52,13 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Hash Table or Array

First, we use a hash table or an array $s$ of length $1001$ to record the elements that appear in the array $nums1$. Then, we iterate through each element in the array $nums2$. If an element $x$ is in $s$, we add $x$ to the answer and remove $x$ from $s$.

After the iteration is finished, we return the answer array.

The time complexity is $O(n+m)$, and the space complexity is $O(n)$. Here, $n$ and $m$ are the lengths of the arrays $nums1$ and $nums2$, respectively.

<!-- tabs:start -->

Expand Down Expand Up @@ -126,6 +132,15 @@ func intersection(nums1 []int, nums2 []int) (ans []int) {
}
```

#### TypeScript

```ts
function intersection(nums1: number[], nums2: number[]): number[] {
const s = new Set(nums1);
return [...new Set(nums2.filter(x => s.has(x)))];
}
```

#### JavaScript

```js
Expand All @@ -135,18 +150,8 @@ func intersection(nums1 []int, nums2 []int) (ans []int) {
* @return {number[]}
*/
var intersection = function (nums1, nums2) {
const s = Array(1001).fill(false);
for (const x of nums1) {
s[x] = true;
}
const ans = [];
for (const x of nums2) {
if (s[x]) {
ans.push(x);
s[x] = false;
}
}
return ans;
const s = new Set(nums1);
return [...new Set(nums2.filter(x => s.has(x)))];
};
```

Expand All @@ -155,15 +160,12 @@ var intersection = function (nums1, nums2) {
```cs
public class Solution {
public int[] Intersection(int[] nums1, int[] nums2) {
List<int> result = new List<int>();
HashSet<int> arr1 = new(nums1);
HashSet<int> arr2 = new(nums2);
foreach (int x in arr1) {
if (arr2.Contains(x)) {
result.Add(x);
}
}
return result.ToArray();
HashSet<int> s1 = new HashSet<int>(nums1);
HashSet<int> s2 = new HashSet<int>(nums2);
s1.IntersectWith(s2);
int[] ans = new int[s1.Count];
s1.CopyTo(ans);
return ans;
}
}
```
Expand All @@ -178,18 +180,10 @@ class Solution {
* @return Integer[]
*/
function intersection($nums1, $nums2) {
$rs = [];
$set1 = array_values(array_unique($nums1));
$set2 = array_values(array_unique($nums2));
for ($i = 0; $i < count($set1); $i++) {
$hashmap[$set1[$i]] = 1;
}
for ($j = 0; $j < count($set2); $j++) {
if ($hashmap[$set2[$j]]) {
array_push($rs, $set2[$j]);
}
}
return $rs;
$s1 = array_unique($nums1);
$s2 = array_unique($nums2);
$ans = array_intersect($s1, $s2);
return array_values($ans);
}
}
```
Expand All @@ -198,27 +192,4 @@ class Solution {

<!-- solution:end -->

<!-- solution:start -->

### Solution 2

<!-- tabs:start -->

#### JavaScript

```js
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number[]}
*/
var intersection = function (nums1, nums2) {
return Array.from(new Set(nums1)).filter(num => new Set(nums2).has(num));
};
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
14 changes: 2 additions & 12 deletions solution/0300-0399/0349.Intersection of Two Arrays/Solution.js
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Expand Up @@ -4,16 +4,6 @@
* @return {number[]}
*/
var intersection = function (nums1, nums2) {
const s = Array(1001).fill(false);
for (const x of nums1) {
s[x] = true;
}
const ans = [];
for (const x of nums2) {
if (s[x]) {
ans.push(x);
s[x] = false;
}
}
return ans;
const s = new Set(nums1);
return [...new Set(nums2.filter(x => s.has(x)))];
};
18 changes: 5 additions & 13 deletions solution/0300-0399/0349.Intersection of Two Arrays/Solution.php
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Expand Up @@ -5,17 +5,9 @@ class Solution {
* @return Integer[]
*/
function intersection($nums1, $nums2) {
$rs = [];
$set1 = array_values(array_unique($nums1));
$set2 = array_values(array_unique($nums2));
for ($i = 0; $i < count($set1); $i++) {
$hashmap[$set1[$i]] = 1;
}
for ($j = 0; $j < count($set2); $j++) {
if ($hashmap[$set2[$j]]) {
array_push($rs, $set2[$j]);
}
}
return $rs;
$s1 = array_unique($nums1);
$s2 = array_unique($nums2);
$ans = array_intersect($s1, $s2);
return array_values($ans);
}
}
}
4 changes: 4 additions & 0 deletions solution/0300-0399/0349.Intersection of Two Arrays/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,4 @@
function intersection(nums1: number[], nums2: number[]): number[] {
const s = new Set(nums1);
return [...new Set(nums2.filter(x => s.has(x)))];
}
8 changes: 0 additions & 8 deletions solution/0300-0399/0349.Intersection of Two Arrays/Solution2.js
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This file was deleted.

17 changes: 7 additions & 10 deletions solution/0300-0399/0349.Intersection of Two Arrays/solution.cs
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Original file line number Diff line number Diff line change
@@ -1,13 +1,10 @@
public class Solution {
public int[] Intersection(int[] nums1, int[] nums2) {
List<int> result = new List<int>();
HashSet<int> arr1 = new(nums1);
HashSet<int> arr2 = new(nums2);
foreach (int x in arr1) {
if (arr2.Contains(x)) {
result.Add(x);
}
}
return result.ToArray();
HashSet<int> s1 = new HashSet<int>(nums1);
HashSet<int> s2 = new HashSet<int>(nums2);
s1.IntersectWith(s2);
int[] ans = new int[s1.Count];
s1.CopyTo(ans);
return ans;
}
}
}
31 changes: 23 additions & 8 deletions solution/2200-2299/2285.Maximum Total Importance of Roads/README.md
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Expand Up @@ -83,9 +83,9 @@ tags:

### 方法一：贪心 + 排序

考虑每个城市对所有道路的总重要性的贡献度，按贡献度从小到大排序，为城市依次分配 $[1, 2, ..., n]$。
我们考虑每个城市对所有道路的总重要性的贡献度，记录在数组 $\text{deg}$ 中。然后将 $\text{deg}$ 按贡献度从小到大排序，为城市依次分配 $[1, 2, ..., n]$。

时间复杂度 $O(n \tiems \log n)$，其中 $n$ 表示城市数目。
时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -135,7 +135,9 @@ public:
}
sort(deg.begin(), deg.end());
long long ans = 0;
for (int i = 0; i < n; ++i) ans += 1ll * (i + 1) * deg[i];
for (int i = 0; i < n; ++i) {
ans += (i + 1LL) * deg[i];
}
return ans;
}
};
Expand All @@ -144,18 +146,31 @@ public:
#### Go

```go
func maximumImportance(n int, roads [][]int) int64 {
func maximumImportance(n int, roads [][]int) (ans int64) {
deg := make([]int, n)
for _, r := range roads {
deg[r[0]]++
deg[r[1]]++
}
sort.Ints(deg)
var ans int64
for i := 0; i < n; i++ {
ans += int64((i + 1) * deg[i])
for i, x := range deg {
ans += int64(x) * int64(i+1)
}
return ans
return
}
```

#### TypeScript

```ts
function maximumImportance(n: number, roads: number[][]): number {
const deg: number[] = Array(n).fill(0);
for (const [a, b] of roads) {
++deg[a];
++deg[b];
}
deg.sort((a, b) => a - b);
return deg.reduce((acc, cur, idx) => acc + (idx + 1) * cur, 0);
}
```

Expand Down
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