算法专题十五：优先级队列-优快云博客

最后一块石头的重量

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> queue=new PriorityQueue<>((a,b)->b-a);
        for(int x:stones){
            queue.offer(x);
        }
        while(queue.size()>1){
            int val1=queue.poll();
            int val2=queue.poll();
            if(val1>val2){
                queue.offer(val1-val2);
            }
        }

        return queue.isEmpty()?0:queue.peek();
    }
}

数据流的第K大元素

703. 数据流中的第 K 大元素 - 力扣（LeetCode）

class KthLargest {
    PriorityQueue<Integer> heap;
    int k1;
    public KthLargest(int k, int[] nums) {
        k1=k;
        heap=new PriorityQueue<>();
        for(int x:nums){
            heap.offer(x);
            if(heap.size()>k1){
                heap.poll();
            }
        }

    }
    
    public int add(int val) {
        heap.offer(val);
        if(heap.size()>k1){
            heap.poll();
        }
        return heap.peek();
    }
}

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

前K个高频单词

692. 前K个高频单词 - 力扣（LeetCode）

在单词出现的次数相同的时候，我们需要比较他们的字典序

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String,Integer> hash=new HashMap<>();
        for(String s:words){
            hash.put(s,hash.getOrDefault(s,0)+1);
        }

        PriorityQueue<Pair<String,Integer>> heap=new PriorityQueue<>(
            (a,b)->{
                if(a.getValue().equals(b.getValue())){
                    return b.getKey().compareTo(a.getKey());
                }
                return a.getValue()-b.getValue();
            }
        );
        for(Map.Entry<String,Integer> e:hash.entrySet()){
             heap.offer(new Pair<>(e.getKey(), e.getValue()));
            if(heap.size()>k){
                heap.poll();
            }
        }

        List<String> ret=new ArrayList<>();
        while(!heap.isEmpty()){
            ret.add(heap.poll().getKey());

        }
        Collections.reverse(ret);
        return ret;

    }
}

数据流的中位数

295. 数据流的中位数 - 力扣（LeetCode）

创建一个大根堆，一个小根堆

class MedianFinder {
    PriorityQueue<Integer> left;
    PriorityQueue<Integer> right;

    public MedianFinder() {
        left=new PriorityQueue<>((a,b)->b-a);
        right=new PriorityQueue<>((a,b)->a-b);

    }
    
    public void addNum(int num) {
        // 长度相等
        if(left.size()==right.size()){
            if(left.isEmpty() || num<=left.peek()){
                left.offer(num);
            }else{
                right.offer(num);
                left.offer(right.poll());
            }

       }else{  //  长度不相等                         
            if(num<=left.peek()){
                left.offer(num);
                right.offer(left.poll());
            }else{
                right.offer(num);
            }

        }
    }
    
    public double findMedian() {
        if(left.size()==right.size()){
            return (left.peek()+right.peek())/2.0;
        }else{
            return left.peek();
        }
    }
}

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder obj = new MedianFinder();
 * obj.addNum(num);
 * double param_2 = obj.findMedian();
 */