feat: add solutions to lcci question: 17.16.The Masseuse

添加《程序员面试金典》题解：面试题 17.16. 按摩师

Author: yanglbme

lcci/17.16.The Masseuse/README.md

@@ -32,20 +32,49 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
+动态规划求解。
 
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def massage(self, nums: List[int]) -> int:
+        if not nums:
+            return 0
+        n = len(nums)
+        if n < 2:
+            return nums[0]
+        dp = [0 for _ in range(n)]
+        dp[0], dp[1] = nums[0], max(nums[0], nums[1])
+        for i in range(2, n):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+        return dp[n - 1]
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int massage(int[] nums) {
+        if (nums == null) {
+            return 0;
+        }
+        int n = nums.length;
+        if (n < 2) {
+            return n == 0 ? 0 : nums[0];
+        }
+        int[] dp = new int[n];
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+        for (int i = 2; i < n; ++i) {
+            dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+        return dp[n - 1];
+    }
+}
 ```
 
 ### ...

lcci/17.16.The Masseuse/README_EN.md

@@ -1,54 +1,111 @@
-# [17.16. The Masseuse](https://leetcode-cn.com/problems/the-masseuse-lcci)
-
-## Description
-<p>A popular masseuse receives a sequence of back-to-back appointment requests and is debating which ones to accept. She needs a break between appointments and therefore she cannot accept any adjacent requests. Given a sequence of back-to-back appoint&shy; ment requests, find the optimal (highest total booked minutes) set the masseuse can honor. Return the number of minutes.</p>
-
-<p><b>Note:&nbsp;</b>This problem is slightly different from the original one in the book.</p>
-
-<p>&nbsp;</p>
-
-<p><strong>Example 1: </strong></p>
-
-<pre>
-<strong>Input: </strong> [1,2,3,1]
-<strong>Output: </strong> 4
-<strong>Explanation: </strong> Accept request 1 and 3, total minutes = 1 + 3 = 4
-</pre>
-
-<p><strong>Example 2: </strong></p>
-
-<pre>
-<strong>Input: </strong> [2,7,9,3,1]
-<strong>Output: </strong> 12
-<strong>Explanation: </strong> Accept request 1, 3 and 5, total minutes = 2 + 9 + 1 = 12
-</pre>
-
-<p><strong>Example 3: </strong></p>
-
-<pre>
-<strong>Input: </strong> [2,1,4,5,3,1,1,3]
-<strong>Output: </strong> 12
-<strong>Explanation: </strong> Accept request 1, 3, 5 and 8, total minutes = 2 + 4 + 3 + 3 = 12
-</pre>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [17.16. The Masseuse](https://leetcode-cn.com/problems/the-masseuse-lcci)
+
+## Description
+<p>A popular masseuse receives a sequence of back-to-back appointment requests and is debating which ones to accept. She needs a break between appointments and therefore she cannot accept any adjacent requests. Given a sequence of back-to-back appoint&shy; ment requests, find the optimal (highest total booked minutes) set the masseuse can honor. Return the number of minutes.</p>
+
+
+
+<p><b>Note:&nbsp;</b>This problem is slightly different from the original one in the book.</p>
+
+
+
+<p>&nbsp;</p>
+
+
+
+<p><strong>Example 1: </strong></p>
+
+
+
+<pre>
+
+<strong>Input: </strong> [1,2,3,1]
+
+<strong>Output: </strong> 4
+
+<strong>Explanation: </strong> Accept request 1 and 3, total minutes = 1 + 3 = 4
+
+</pre>
+
+
+
+<p><strong>Example 2: </strong></p>
+
+
+
+<pre>
+
+<strong>Input: </strong> [2,7,9,3,1]
+
+<strong>Output: </strong> 12
+
+<strong>Explanation: </strong> Accept request 1, 3 and 5, total minutes = 2 + 9 + 1 = 12
+
+</pre>
+
+
+
+<p><strong>Example 3: </strong></p>
+
+
+
+<pre>
+
+<strong>Input: </strong> [2,1,4,5,3,1,1,3]
+
+<strong>Output: </strong> 12
+
+<strong>Explanation: </strong> Accept request 1, 3, 5 and 8, total minutes = 2 + 4 + 3 + 3 = 12
+
+</pre>
+
+
+
+
+## Solutions
+
+
+### Python3
+
+```python
+class Solution:
+    def massage(self, nums: List[int]) -> int:
+        if not nums:
+            return 0
+        n = len(nums)
+        if n < 2:
+            return nums[0]
+        dp = [0 for _ in range(n)]
+        dp[0], dp[1] = nums[0], max(nums[0], nums[1])
+        for i in range(2, n):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+        return dp[n - 1]
+```
+
+### Java
+
+```java
+class Solution {
+    public int massage(int[] nums) {
+        if (nums == null) {
+            return 0;
+        }
+        int n = nums.length;
+        if (n < 2) {
+            return n == 0 ? 0 : nums[0];
+        }
+        int[] dp = new int[n];
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+        for (int i = 2; i < n; ++i) {
+            dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+        return dp[n - 1];
+    }
+}
+```
+
+### ...
+```
+
+```

lcci/17.16.The Masseuse/Solution.java

@@ -0,0 +1,18 @@
+class Solution {
+    public int massage(int[] nums) {
+        if (nums == null) {
+            return 0;
+        }
+        int n = nums.length;
+        if (n < 2) {
+            return n == 0 ? 0 : nums[0];
+        }
+        int[] dp = new int[n];
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+        for (int i = 2; i < n; ++i) {
+            dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+        return dp[n - 1];
+    }
+}

lcci/17.16.The Masseuse/Solution.py

@@ -0,0 +1,12 @@
+class Solution:
+    def massage(self, nums: List[int]) -> int:
+        if not nums:
+            return 0
+        n = len(nums)
+        if n < 2:
+            return nums[0]
+        dp = [0 for _ in range(n)]
+        dp[0], dp[1] = nums[0], max(nums[0], nums[1])
+        for i in range(2, n):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+        return dp[n - 1]