703. Kth Largest Element in a Stream

 

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.

 

这道题让我们在数据流中求第K大的元素,跟之前那道Kth Largest Element in an Array很类似,但不同的是,那道题的数组是确定的,不会再增加元素,这样确定第K大的数字就比较简单。而这道题的数组是不断在变大的,所以每次第K大的数字都在不停的变化。那么我们其实只关心前K大个数字就可以了,所以我们可以使用一个最小堆来保存前K个数字,当再加入新数字后,最小堆会自动排序,然后把排序后的最小的那个数字去除,则堆中还是K个数字,返回的时候只需返回堆顶元素即可,参见代码如下:

 

解法一:

class KthLargest {
public:
    KthLargest(int k, vector<int> nums) {
        for (int num : nums) {
            q.push(num);
            if (q.size() > k) q.pop();
        }
        K = k;
    }
    
    int add(int val) {
        q.push(val);
        if (q.size() > K) q.pop();
        return q.top();
    }

private:
    priority_queue<int, vector<int>, greater<int>> q;
    int K;
};

 

我们也可以使用multiset来做,利用其可重复,且自动排序的功能,这样也可以达到最小堆的效果,参见代码如下:

 

解法二:

class KthLargest {
public:
    KthLargest(int k, vector<int> nums) {
        for (int num : nums) {
            st.insert(num);
            if (st.size() > k) st.erase(st.begin());
        }
        K = k;
    }
    
    int add(int val) {
        st.insert(val);
        if (st.size() > K) st.erase(st.begin());
        return *st.begin();
    }

private:
    multiset<int> st;
    int K;
};

 

类似题目:

Kth Largest Element in an Array

 

参考资料:

https://leetcode.com/problems/kth-largest-element-in-a-stream

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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