Problem

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given[3,2,1,5,6,4]and k = 2, return 5.

Note:
You may assume k is always valid, 1 ? k ? array's length.

思路一(排序取倒数第k个)

• 时间复杂度: O(nlogn)
• 空间复杂度: O(1)

思路二(minHeap)

• 时间复杂度: O(nlogk)

• 空间复杂度: O(k)

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

思路三(quick select)

• 时间复杂度: O(n)

• 空间复杂度: O(1)

    public int findKthLargest(int[] nums, int k) {
        return quickSelect(nums, 0, nums.length - 1, k);
    }

    private int quickSelect(int[] nums, int start, int end, int k) {
        if (start >= end) {
            return nums[start];
        }
        int left = start;
        int right = end;
        int pivot = nums[(left + right) / 2];

        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;
            }
        }
        if (start + k - 1 <= right) {
            return quickSelect(nums, start, right, k);
        }
        if (start + k - 1 >= left) {
            return quickSelect(nums, left, end, k - (left - start));
        }
        return nums[right + 1];
    }