Problem

Given an integer array, find the top_k_largest numbers in it.

Example

Given[3,10,1000,-99,4,100]andk=3.
Return[1000, 100, 10].

思路一(heap)

    public int[] topk(int[] nums, int k) {
        int[] res = new int[k];
        PriorityQueue<Integer> heap = new PriorityQueue<>();
        for (int n: nums) {
            heap.offer(n);
            if (heap.size() > k) {
                heap.poll();
            }
        }
        for (int i = heap.size() - 1; i >= 0; i--) {
            res[i] = heap.poll();
        }
        return res;
    }

思路二(quick sort)

• quicksort, 从大到小排序, 只排列从头到第k个元素, 然后startIndex >=k 时停止排序
• 时间复杂度: O(klogk) 和 O(nlogn) 之间

    public int[] topk(int[] nums, int k) {
        int[] res = new int[k];
        if (nums == null || nums.length == 0) {
            return res;
        }
        quicksort(nums, 0, nums.length - 1, k);
        for (int i = 0; i < k; i++) {
            res[i] = nums[i];
            System.out.println(res[i]);
        }
        return res;
    }

    private void quicksort(int[] nums, int start, int end, int k) {
        if (start >= k || start >= end) {
            return;
        }
        int pivot = nums[start + (end - start) / 2];
        int left = start;
        int right = end;
        while (left <= right) {
            while (left <= right && nums[left] > pivot) {
                left++;
            }
            while (left <= right && nums[right] < pivot) {
                right--;
            }
            if (left <= right) {
                int temp = nums[left];
                nums[left] = nums[right];
                nums[right] = temp;
                left++;
                right--;
            }
        }
        quicksort(nums, start, right, k);
        quicksort(nums, left, end, k);
    }