题目

法1：小根堆

最大的K个元素 => 小根堆（类似上窄下宽的梯形）
最小的K个元素 => 大根堆（类似倒三角形）
必须掌握！！！

class Solution {public int findKthLargest(int[] nums, int k) {PriorityQueue<Integer> q = new PriorityQueue<>();for (int i = 0; i < k; ++i) {q.offer(nums[i]);}for (int i = k; i < nums.length; ++i) {if (nums[i] > q.peek()) {q.poll();q.offer(nums[i]);}}return q.peek();}
}

法2：基于快排

class Solution {public int findKthLargest(int[] nums, int k) {List<Integer> list = new ArrayList<>();for (int i : nums) {list.add(i);}return quickSelect(list, k);}public int quickSelect(List<Integer> list, int k) {int n = list.size();Random rand = new Random();int rInx = rand.nextInt(n); // 生成0 ~ n-1之间的整数List<Integer> big = new ArrayList<>();List<Integer> same = new ArrayList<>();List<Integer> small = new ArrayList<>();for (int i = 0; i < n; ++i) {if (list.get(i) > list.get(rInx)) {big.add(list.get(i));} else if (list.get(i) == list.get(rInx)) {same.add(list.get(i));} else {small.add(list.get(i));}}if (big.size() >= k) {return quickSelect(big, k);} else if (k > (big.size() + same.size())) {return quickSelect(small, k - (big.size() + same.size()));} else {return list.get(rInx);}}
}