【LeetCode 0703】【设计】【优先队列】返回数据流中第K大的元素

703. Kth Largest Element in a Stream

Design a class to find the k^th largest element in a stream. Note that it is the k^th largest element in the sorted order, not the k^th distinct element.

Implement KthLargest class:

• KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.
• int add(int val) Appends the integer val to the stream and returns the element representing the k^th largest element in the stream.

Example 1:

**Input**
["KthLargest", "add", "add", "add", "add", "add"]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
**Output**
[null, 4, 5, 5, 8, 8]

**Explanation**
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3);   // return 4
kthLargest.add(5);   // return 5
kthLargest.add(10);  // return 5
kthLargest.add(9);   // return 8
kthLargest.add(4);   // return 8

Constraints:

• 1 <= k <= 10^4
• 0 <= nums.length <= 10^4
• -10^4 <= nums[i] <= 10^4
• -10^4 <= val <= 10^4
• At most 10^4 calls will be made to add.
• It is guaranteed that there will be at least k elements in the array when you search for the k^th element.

Idea

1.基于数组实现/维护一个大小size=k的小堆
2.具体处理
当读入的元素个数不大于k个，直接入堆
当读入元素超过k个，则比较堆顶元素是否比下一个读入数要小
* 如果读入的元素比堆顶元素小，则忽略
* 反之，将读入的元素入堆，并调整/挤掉堆顶元素
最后堆顶元素即为第K大的元素

JavaScript Solution

/**
 * @param {number} k
 * @param {number[]} nums
 */
var KthLargest = function(k, nums) {
    
};

/** 
 * @param {number} val
 * @return {number}
 */
KthLargest.prototype.add = function(val) {
    
};

/** 
 * Your KthLargest object will be instantiated and called as such:
 * var obj = new KthLargest(k, nums)
 * var param_1 = obj.add(val)
 */