feat: add solutions to lc problem: No.2611

No.2611.Mice and Cheese

Author: yanglbme

solution/2600-2699/2611.Mice and Cheese/README.md

@@ -55,6 +55,14 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：贪心 + 排序**
+
+我们可以先将所有奶酪分给第二只老鼠，接下来，考虑将其中 $k$ 块奶酪分给第一只老鼠，那么我们应该如何选择这 $k$ 块奶酪呢？显然，将第 $i$ 块奶酪从第二只老鼠分给第一只老鼠，得分的变化量为 $reward1[i] - reward2[i]$，我们希望这个变化量尽可能大，这样才能使得总得分最大。
+
+因此，我们将奶酪按照 `reward1[i] - reward2[i]` 从大到小排序，前 $k$ 块奶酪由第一只老鼠吃掉，剩下的奶酪由第二只老鼠吃掉，即可得到最大得分。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为奶酪的数量。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -70,6 +78,15 @@ class Solution:
         return sum(reward1[i] for i in idx[:k]) + sum(reward2[i] for i in idx[k:])
 ```
 
+```python
+class Solution:
+    def miceAndCheese(self, reward1: List[int], reward2: List[int], k: int) -> int:
+        for i, x in enumerate(reward2):
+            reward1[i] -= x
+        reward1.sort(reverse=True)
+        return sum(reward2) + sum(reward1[:k])
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -95,6 +112,24 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    public int miceAndCheese(int[] reward1, int[] reward2, int k) {
+        int ans = 0;
+        int n = reward1.length;
+        for (int i = 0; i < n; ++i) {
+            ans += reward2[i];
+            reward1[i] -= reward2[i];
+        }
+        Arrays.sort(reward1);
+        for (int i = 0; i < k; ++i) {
+            ans += reward1[n - i - 1];
+        }
+        return ans;
+    }
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -117,6 +152,23 @@ public:
 };
 ```
 
+```cpp
+class Solution {
+public:
+    int miceAndCheese(vector<int>& reward1, vector<int>& reward2, int k) {
+        int n = reward1.size();
+        int ans = 0;
+        for (int i = 0; i < n; ++i) {
+            ans += reward2[i];
+            reward1[i] -= reward2[i];
+        }
+        sort(reward1.rbegin(), reward1.rend());
+        ans += accumulate(reward1.begin(), reward1.begin() + k, 0);
+        return ans;
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -140,6 +192,21 @@ func miceAndCheese(reward1 []int, reward2 []int, k int) (ans int) {
 }
 ```
 
+```go
+func miceAndCheese(reward1 []int, reward2 []int, k int) (ans int) {
+	for i, x := range reward2 {
+		ans += x
+		reward1[i] -= x
+	}
+	sort.Ints(reward1)
+	n := len(reward1)
+	for i := 0; i < k; i++ {
+		ans += reward1[n-i-1]
+	}
+	return
+}
+```
+
 ### **...**
 
 ```

solution/2600-2699/2611.Mice and Cheese/README_EN.md

@@ -62,6 +62,15 @@ class Solution:
         return sum(reward1[i] for i in idx[:k]) + sum(reward2[i] for i in idx[k:])
 ```
 
+```python
+class Solution:
+    def miceAndCheese(self, reward1: List[int], reward2: List[int], k: int) -> int:
+        for i, x in enumerate(reward2):
+            reward1[i] -= x
+        reward1.sort(reverse=True)
+        return sum(reward2) + sum(reward1[:k])
+```
+
 ### **Java**
 
 ```java
@@ -85,6 +94,24 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    public int miceAndCheese(int[] reward1, int[] reward2, int k) {
+        int ans = 0;
+        int n = reward1.length;
+        for (int i = 0; i < n; ++i) {
+            ans += reward2[i];
+            reward1[i] -= reward2[i];
+        }
+        Arrays.sort(reward1);
+        for (int i = 0; i < k; ++i) {
+            ans += reward1[n - i - 1];
+        }
+        return ans;
+    }
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -107,6 +134,23 @@ public:
 };
 ```
 
+```cpp
+class Solution {
+public:
+    int miceAndCheese(vector<int>& reward1, vector<int>& reward2, int k) {
+        int n = reward1.size();
+        int ans = 0;
+        for (int i = 0; i < n; ++i) {
+            ans += reward2[i];
+            reward1[i] -= reward2[i];
+        }
+        sort(reward1.rbegin(), reward1.rend());
+        ans += accumulate(reward1.begin(), reward1.begin() + k, 0);
+        return ans;
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -130,6 +174,21 @@ func miceAndCheese(reward1 []int, reward2 []int, k int) (ans int) {
 }
 ```
 
+```go
+func miceAndCheese(reward1 []int, reward2 []int, k int) (ans int) {
+	for i, x := range reward2 {
+		ans += x
+		reward1[i] -= x
+	}
+	sort.Ints(reward1)
+	n := len(reward1)
+	for i := 0; i < k; i++ {
+		ans += reward1[n-i-1]
+	}
+	return
+}
+```
+
 ### **...**
 
 ```