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[LintCode] Kth Largest Element [PriorityQueue]

Hwg / 3066人閱讀

摘要:可以不要用太簡單的方法。先把它裝滿,再和隊列頂端的數(shù)字比較,大的就替換掉,小的就。遍歷完所有元素之后,頂部的數(shù)就是第大的數(shù)。

Problem

Find K-th largest element in an array.

Example

In array [9,3,2,4,8], the 3rd largest element is 4.
In array [1,2,3,4,5], the 1st largest element is 5, 2nd largest element is 4, 3rd largest element is 3 and etc.

Note

可以不要用太簡單的方法。
什么和Arrays.sort()最接近?PriorityQueue.
An unbounded priority queue based on a priority heap. The elements of the priority queue are ordered according to their natural ordering, or by a Comparator provided at queue construction time, depending on which constructor is used.
A priority queue does not permit null elements.
A priority queue relying on natural ordering also does not permit insertion of non-comparable objects (doing so may result in ClassCastException).

做一個大小為k的PriorityQueue,peek()到的最頂端元素是隊列中最小的元素,那么PQ就像一個自動過濾掉比頂端元素更小元素并把內(nèi)部所有元素進行排序的容器。先把它裝滿,再和隊列頂端的數(shù)字比較,大的就替換掉,小的就continue。遍歷完所有元素之后,頂部的數(shù)就是第k大的數(shù)。
熟悉PriorityQueue的操作,.add(), .peek(), .remove().

Solution

1.

class Solution {
    public int kthLargestElement(int k, int[] nums) {
        Arrays.sort(nums);
        int len = nums.length;
        return nums[len - k];
    }
}

2.

class Solution {
    public int kthLargestElement(int k, int[] nums) {
        PriorityQueue queue = new PriorityQueue(k);
        for (int num : nums) {
            if (queue.size() < k) {
                queue.add(num);
            } else if (queue.peek() < num) {
                queue.remove();
                queue.add(num);
            }
        }
        return queue.peek();
    }
}

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