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);
        }
    }
}