feat: add solutions to lc problems: No.0349,2285

* No.0349.Intersection of Two Arrays
* No.2285.Maximum Total Importance of Roads

Author: yanglbme

solution/0300-0399/0349.Intersection of Two Arrays/README.md

@@ -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
@@ -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)))];
 };
 ```
 
@@ -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;
     }
 }
 ```
@@ -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);
     }
 }
 ```
@@ -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 -->

solution/0300-0399/0349.Intersection of Two Arrays/README_EN.md

@@ -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 -->
 
@@ -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
@@ -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)))];
 };
 ```
 
@@ -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;
     }
 }
 ```
@@ -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);
     }
 }
 ```
@@ -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 -->

solution/0300-0399/0349.Intersection of Two Arrays/Solution.js

@@ -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)))];
 };

solution/0300-0399/0349.Intersection of Two Arrays/Solution.php

@@ -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);
     }
-}
+}

solution/0300-0399/0349.Intersection of Two Arrays/Solution.ts

@@ -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)))];
+}

solution/0300-0399/0349.Intersection of Two Arrays/Solution2.js

@@ -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;
     }
-}
+}

solution/0300-0399/0349.Intersection of Two Arrays/solution.cs

@@ -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 -->
 
@@ -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;
     }
 };
@@ -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);
 }
 ```