feat: add solutions to lc problems: No.2956,2957

solution/2900-2999/2956.Find Common Elements Between Two Arrays/README.md

@@ -52,34 +52,128 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：哈希表或数组**
+
+我们可以用两个哈希表或数组 $s1$ 和 $s2$ 分别记录两个数组中出现的元素。
+
+接下来，我们创建一个长度为 $2$ 的数组 $ans$，其中 $ans[0]$ 表示 $nums1$ 中出现在 $s2$ 中的元素个数，$ans[1]$ 表示 $nums2$ 中出现在 $s1$ 中的元素个数。
+
+然后，我们遍历数组 $nums1$ 中的每个元素 $x$，如果 $x$ 在 $s2$ 中出现过，则将 $ans[0]$ 加一。接着，我们遍历数组 $nums2$ 中的每个元素 $x$，如果 $x$ 在 $s1$ 中出现过，则将 $ans[1]$ 加一。
+
+最后，我们返回数组 $ans$ 即可。
+
+时间复杂度 $O(n + m)$，空间复杂度 $O(n + m)$。其中 $n$ 和 $m$ 分别是数组 $nums1$ 和 $nums2$ 的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        s1, s2 = set(nums1), set(nums2)
+        return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int[] findIntersectionValues(int[] nums1, int[] nums2) {
+        int[] s1 = new int[101];
+        int[] s2 = new int[101];
+        for (int x : nums1) {
+            s1[x] = 1;
+        }
+        for (int x : nums2) {
+            s2[x] = 1;
+        }
+        int[] ans = new int[2];
+        for (int x : nums1) {
+            ans[0] += s2[x];
+        }
+        for (int x : nums2) {
+            ans[1] += s1[x];
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
-
+class Solution {
+public:
+    vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
+        int s1[101]{};
+        int s2[101]{};
+        for (int& x : nums1) {
+            s1[x] = 1;
+        }
+        for (int& x : nums2) {
+            s2[x] = 1;
+        }
+        vector<int> ans(2);
+        for (int& x : nums1) {
+            ans[0] += s2[x];
+        }
+        for (int& x : nums2) {
+            ans[1] += s1[x];
+        }
+        return ans;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+func findIntersectionValues(nums1 []int, nums2 []int) []int {
+	s1 := [101]int{}
+	s2 := [101]int{}
+	for _, x := range nums1 {
+		s1[x] = 1
+	}
+	for _, x := range nums2 {
+		s2[x] = 1
+	}
+	ans := make([]int, 2)
+	for _, x := range nums1 {
+		ans[0] += s2[x]
+	}
+	for _, x := range nums2 {
+		ans[1] += s1[x]
+	}
+	return ans
+}
+```
 
+### **TypeScript**
+
+```ts
+function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
+    const s1: number[] = Array(101).fill(0);
+    const s2: number[] = Array(101).fill(0);
+    for (const x of nums1) {
+        s1[x] = 1;
+    }
+    for (const x of nums2) {
+        s2[x] = 1;
+    }
+    const ans: number[] = Array(2).fill(0);
+    for (const x of nums1) {
+        ans[0] += s2[x];
+    }
+    for (const x of nums2) {
+        ans[1] += s1[x];
+    }
+    return ans;
+}
 ```
 
 ### **...**

solution/2900-2999/2956.Find Common Elements Between Two Arrays/README_EN.md

@@ -46,30 +46,124 @@
 
 ## Solutions
 
+**Solution 1: Hash Table or Array**
+
+We can use two hash tables or arrays $s1$ and $s2$ to record the elements that appear in the two arrays respectively.
+
+Next, we create an array $ans$ of length $2$, where $ans[0]$ represents the number of elements in $nums1$ that appear in $s2$, and $ans[1]$ represents the number of elements in $nums2$ that appear in $s1$.
+
+Then, we traverse each element $x$ in the array $nums1$. If $x$ has appeared in $s2$, we increment $ans[0]$. After that, we traverse each element $x$ in the array $nums2$. If $x$ has appeared in $s1$, we increment $ans[1]$.
+
+Finally, we return the array $ans$.
+
+The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the lengths of the arrays $nums1$ and $nums2$ respectively.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        s1, s2 = set(nums1), set(nums2)
+        return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int[] findIntersectionValues(int[] nums1, int[] nums2) {
+        int[] s1 = new int[101];
+        int[] s2 = new int[101];
+        for (int x : nums1) {
+            s1[x] = 1;
+        }
+        for (int x : nums2) {
+            s2[x] = 1;
+        }
+        int[] ans = new int[2];
+        for (int x : nums1) {
+            ans[0] += s2[x];
+        }
+        for (int x : nums2) {
+            ans[1] += s1[x];
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
-
+class Solution {
+public:
+    vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
+        int s1[101]{};
+        int s2[101]{};
+        for (int& x : nums1) {
+            s1[x] = 1;
+        }
+        for (int& x : nums2) {
+            s2[x] = 1;
+        }
+        vector<int> ans(2);
+        for (int& x : nums1) {
+            ans[0] += s2[x];
+        }
+        for (int& x : nums2) {
+            ans[1] += s1[x];
+        }
+        return ans;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+func findIntersectionValues(nums1 []int, nums2 []int) []int {
+	s1 := [101]int{}
+	s2 := [101]int{}
+	for _, x := range nums1 {
+		s1[x] = 1
+	}
+	for _, x := range nums2 {
+		s2[x] = 1
+	}
+	ans := make([]int, 2)
+	for _, x := range nums1 {
+		ans[0] += s2[x]
+	}
+	for _, x := range nums2 {
+		ans[1] += s1[x]
+	}
+	return ans
+}
+```
 
+### **TypeScript**
+
+```ts
+function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
+    const s1: number[] = Array(101).fill(0);
+    const s2: number[] = Array(101).fill(0);
+    for (const x of nums1) {
+        s1[x] = 1;
+    }
+    for (const x of nums2) {
+        s2[x] = 1;
+    }
+    const ans: number[] = Array(2).fill(0);
+    for (const x of nums1) {
+        ans[0] += s2[x];
+    }
+    for (const x of nums2) {
+        ans[1] += s1[x];
+    }
+    return ans;
+}
 ```
 
 ### **...**

solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.cpp

@@ -0,0 +1,21 @@
+class Solution {
+public:
+    vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
+        int s1[101]{};
+        int s2[101]{};
+        for (int& x : nums1) {
+            s1[x] = 1;
+        }
+        for (int& x : nums2) {
+            s2[x] = 1;
+        }
+        vector<int> ans(2);
+        for (int& x : nums1) {
+            ans[0] += s2[x];
+        }
+        for (int& x : nums2) {
+            ans[1] += s1[x];
+        }
+        return ans;
+    }
+};

solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.go

@@ -0,0 +1,18 @@
+func findIntersectionValues(nums1 []int, nums2 []int) []int {
+	s1 := [101]int{}
+	s2 := [101]int{}
+	for _, x := range nums1 {
+		s1[x] = 1
+	}
+	for _, x := range nums2 {
+		s2[x] = 1
+	}
+	ans := make([]int, 2)
+	for _, x := range nums1 {
+		ans[0] += s2[x]
+	}
+	for _, x := range nums2 {
+		ans[1] += s1[x]
+	}
+	return ans
+}

solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.java

@@ -0,0 +1,20 @@
+class Solution {
+    public int[] findIntersectionValues(int[] nums1, int[] nums2) {
+        int[] s1 = new int[101];
+        int[] s2 = new int[101];
+        for (int x : nums1) {
+            s1[x] = 1;
+        }
+        for (int x : nums2) {
+            s2[x] = 1;
+        }
+        int[] ans = new int[2];
+        for (int x : nums1) {
+            ans[0] += s2[x];
+        }
+        for (int x : nums2) {
+            ans[1] += s1[x];
+        }
+        return ans;
+    }
+}

solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.py

@@ -0,0 +1,4 @@
+class Solution:
+    def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        s1, s2 = set(nums1), set(nums2)
+        return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]

solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.ts

@@ -0,0 +1,18 @@
+function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
+    const s1: number[] = Array(101).fill(0);
+    const s2: number[] = Array(101).fill(0);
+    for (const x of nums1) {
+        s1[x] = 1;
+    }
+    for (const x of nums2) {
+        s2[x] = 1;
+    }
+    const ans: number[] = Array(2).fill(0);
+    for (const x of nums1) {
+        ans[0] += s2[x];
+    }
+    for (const x of nums2) {
+        ans[1] += s1[x];
+    }
+    return ans;
+}