/***最最最最典型的topK的问题,在这里先用优先队列实现一下子*/public class Solution {public int findKthLargest(int[] nums, int k) {PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>(k, new Comparator<Integer>() {@Overridepublic int compare(Integer o1, Integer o2) {return o1.compareTo(o2);}});for(int i : nums){if(priorityQueue.size() >= k){if(priorityQueue.peek() < i){priorityQueue.poll();priorityQueue.add(i);}}else {priorityQueue.add(i);}}return priorityQueue.poll();}}
