Add solution 703

Author: yanglbme

README.md

@@ -36,6 +36,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 235 | [Lowest Common Ancestor of a Binary Search Tree](https://github.com/doocs/leetcode/tree/master/solution/235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree) | `Tree` |
 | 237 | [Delete Node in a Linked List](https://github.com/doocs/leetcode/tree/master/solution/237.Delete%20Node%20in%20a%20Linked%20List) | `Linked List` |
 | 344 | [Reverse String](https://github.com/doocs/leetcode/tree/master/solution/344.Reverse%20String) | `Two Pointers`, `String` |
+| 703 | [Kth Largest Element in a Stream](%20ttps://github.com/doocs/leetcode/tree/master/solution/703.Kth%20Largest%20Element%20in%20a%20Stream) | `Heap` |
 | 876 | [Middle of the Linked List](https://github.com/doocs/leetcode/tree/master/solution/876.Middle%20of%20the%20Linked%20List) | `Linked List` |
 
 

solution/703.Kth Largest Element in a Stream/README.md

@@ -0,0 +1,77 @@
+## 数据流中的第K大元素
+### 题目描述
+
+设计一个找到数据流中第 K 大元素的类（class）。注意是排序后的第 K 大元素，不是第 K 个不同的元素。
+
+你的 `KthLargest` 类需要一个同时接收整数 `k` 和整数数组`nums` 的构造器，它包含数据流中的初始元素。每次调用 `KthLargest.add`，返回当前数据流中第 K 大的元素。
+
+示例:
+```
+int k = 3;
+int[] arr = [4,5,8,2];
+KthLargest kthLargest = new KthLargest(3, arr);
+kthLargest.add(3);   // returns 4
+kthLargest.add(5);   // returns 5
+kthLargest.add(10);  // returns 5
+kthLargest.add(9);   // returns 8
+kthLargest.add(4);   // returns 8
+```
+
+说明: 
+
+你可以假设 `nums` 的长度≥ `k-1` 且 `k` ≥ 1。
+
+### 解法
+建立一个有 k 个元素的小根堆。add 操作时：
+
+- 若堆元素少于 k 个，直接添加到堆中；
+- 若堆元素有 k 个，判断堆顶元素 peek() 与 val 的大小关系：若 peek() >= val，直接返回堆顶元素；若 peek() < val，弹出堆顶元素，将 val 添加至堆中，然后返回堆顶。
+
+```java
+class KthLargest {
+
+    PriorityQueue<Integer> queue;
+    
+    private int size = 0;
+    
+    public KthLargest(int k, int[] nums) {
+        size = k;
+        queue = new PriorityQueue<>(k, Integer::compareTo);
+        int gap = nums.length - k;
+        if (gap > 0) {
+            for (int i = 0; i < k; ++i) {
+                queue.offer(nums[i]);
+            }
+            for (int i = k; i < nums.length; ++i) {
+                add(nums[i]);
+            }
+        } else {
+            for (int i = 0; i < nums.length; ++i) {
+                queue.offer(nums[i]);
+            }
+        }
+    }
+    
+    public int add(int val) {
+        if (queue.size() < size) {
+            queue.offer(val);
+            return queue.peek();
+        }
+        
+        if (queue.peek() >= val) {
+            return queue.peek();
+        }
+        
+        queue.poll();
+        queue.offer(val);
+        return queue.peek();
+        
+    }
+}
+
+/**
+ * Your KthLargest object will be instantiated and called as such:
+ * KthLargest obj = new KthLargest(k, nums);
+ * int param_1 = obj.add(val);
+ */
+```

solution/703.Kth Largest Element in a Stream/Solution.java

@@ -0,0 +1,46 @@
+class KthLargest {
+
+    PriorityQueue<Integer> queue;
+
+    private int size = 0;
+
+    public KthLargest(int k, int[] nums) {
+        size = k;
+        queue = new PriorityQueue<>(k, Integer::compareTo);
+        int gap = nums.length - k;
+        if (gap > 0) {
+            for (int i = 0; i < k; ++i) {
+                queue.offer(nums[i]);
+            }
+            for (int i = k; i < nums.length; ++i) {
+                add(nums[i]);
+            }
+        } else {
+            for (int i = 0; i < nums.length; ++i) {
+                queue.offer(nums[i]);
+            }
+        }
+    }
+
+    public int add(int val) {
+        if (queue.size() < size) {
+            queue.offer(val);
+            return queue.peek();
+        }
+
+        if (queue.peek() >= val) {
+            return queue.peek();
+        }
+
+        queue.poll();
+        queue.offer(val);
+        return queue.peek();
+
+    }
+}
+
+/**
+ * Your KthLargest object will be instantiated and called as such: 
+ * KthLargest obj = new KthLargest(k, nums); 
+ * int param_1 = obj.add(val);
+ */