Update 150 Java Version with SwitchCase

problems/0150.逆波兰表达式求值.md

@@ -162,6 +162,39 @@ class Solution {
 }
 ```
 
+```Java
+// Switch Case 写法，更简洁一些
+class Solution {
+    public int evalRPN(String[] tokens) {
+        Stack<Integer> stk = new Stack<>();
+        for (String token : tokens) {
+
+            // 运算符Operator + - * /
+            if ("+-*/".contains(token)) {
+                int a = stk.pop(), b = stk.pop();
+                switch (token) {
+                    case "+":
+                        stk.push(a + b);
+                        break;
+                    case "-":
+                        stk.push(b - a);
+                        break;
+                    case "*":
+                        stk.push(a * b);
+                        break;
+                    case "/":
+                        stk.push(b / a);
+                        break;
+                }
+            } else {
+                stk.push(Integer.parseInt(token));
+            }
+        }
+        return stk.pop();
+    }
+}
+```
+
 ### Python3：
 
 ```python

problems/0239.滑动窗口最大值.md

@@ -260,7 +260,9 @@ class Solution {
         return res;
     }
 }
+```
 
+```Java
 //解法二
 //利用双端队列手动实现单调队列
 /**
@@ -296,6 +298,39 @@ class Solution {
 }
 ```
 
+```Java
+//解法三
+//用优先队列 Priority Queue, 时间Time: O(nlogk) | 空间Space: O(k)
+/**
+* 很好理解，PQ 会优先置顶最大值（Java Comparator）
+* 维护window里的 PQ， PQ内部记录 <value, index> pair
+**/
+class Solution {
+    public int[] maxSlidingWindow(int[] nums, int k) {
+        if (nums == null || k <= 0)
+            return new int[0];
+
+        int n = nums.length;
+        int[] res = new int[n - k + 1];
+        PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a, b) -> b[0] - a[0]);
+
+        for (int i = 0; i < n; i++) {
+            maxHeap.offer(new int[] { nums[i], i });
+
+            // 删除窗口之外的元素
+            while (maxHeap.peek()[1] <= i - k) {
+                maxHeap.poll();
+            }
+
+            if (i >= k - 1) {
+                res[i - k + 1] = maxHeap.peek()[0];
+            }
+        }
+        return res;
+    }
+}
+```
+
 ### Python：
 #### 解法一：使用自定义的单调队列类
 ```python

problems/0347.前K个高频元素.md

@@ -213,6 +213,49 @@ class Solution {
         return res;
     }
 }
+```
+Java解法三：Bucket Sort 桶排序
+```java
+/**
+* 利用 桶排序
+* 时间复杂度 O(n)
+* 空间复杂度 O(n)
+**/
+class Solution {
+    public int[] topKFrequent(int[] nums, int k) {
+
+        Map<Integer, Integer> freqMap = new HashMap<>();
+        for (int num : nums) {
+            freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
+        }
+
+        List<Integer>[] bucket = new List[nums.length + 1];
+        for (int i = 0; i <= nums.length; i++) {
+            bucket[i] = new ArrayList<>();
+        }
+
+        for (Map.Entry<Integer, Integer> entry : freqMap.entrySet()) {
+            int num = entry.getKey();
+            int freq = entry.getValue();
+            bucket[freq].add(num);
+        }
+
+        List<Integer> result = new ArrayList<>();
+        for (int i = bucket.length - 1; i >= 0 && result.size() < k; i--) {
+            if (!bucket[i].isEmpty()) {
+                result.addAll(bucket[i]);
+            }
+        }
+
+        int[] res = new int[k];
+        for (int i = 0; i < k; i++) {
+            res[i] = result.get(i);
+        }
+
+        return res;
+    }
+}
+
 ```
 
 ### Python：