703. Kth Largest Element in a Stream

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本文介绍了一种使用最小堆实现的数据结构——KthLargest类，用于高效找到数据流中的第K大元素。该类可在数据流增长时动态维护前K大元素，并快速返回第K大值。

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

Your KthLargest class will have a constructor which accepts an integer k and an integer array nums, which contains initial elements from the stream. For each call to the method KthLargest.add, return the element representing the kth largest element in the stream.

Example:

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
Note:
You may assume that nums’ length ≥ k-1 and k ≥ 1.
该题目要求数据流中的最k大的值，数据流说明了数据是动态增长的。这种问题看到了就想到用堆来存k大的数，这儿我们就要用最小堆。

class KthLargest {
private:
	vector<int> c;
    int K=0;
public:
	KthLargest(int k, vector<int>& nums) {
        K=k;
		make_heap(c.begin(), c.end(), greater<int>());
		for (int i = 0; i < nums.size(); i++)
		{
			if (i < k)
			{
				c.push_back(nums[i]);
				push_heap(c.begin(), c.end(), greater<int>());
                
			}
			else
			{
				int tmp = c.front();
				if (tmp < nums[i])
				{
					c.push_back(nums[i]);
					push_heap(c.begin(), c.end(), greater<int>());
					pop_heap(c.begin(), c.end(), greater<int>());
					c.pop_back();
				}
			}
		}

	}

	int add(int val) {
        if (c.size()<K)
        {
            c.push_back(val);
            push_heap(c.begin(),c.end(),greater<int>());
            return c.front();
        }
		int tmp = c.front();
		if (tmp < val)
		{
			c.push_back(val);
			push_heap(c.begin(), c.end(), greater<int>());
			pop_heap(c.begin(), c.end(), greater<int>());
			c.pop_back();
		}
		return c.front();
	}

};