915. Partition Array into Disjoint Intervals

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最新推荐文章于 2024-10-26 17:25:00 发布

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本文介绍了一种有效的数组分区算法，该算法将数组A分为两个连续的子数组left和right，确保left中的每个元素都小于等于right中的每个元素，同时使left的大小尽可能小。通过示例说明了算法的工作原理，如[5,0,3,8,6]被分为left=[5,0,3]和right=[8,6]。

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Given an array A, partition it into two (contiguous) subarrays left and right so that:

Every element in left is less than or equal to every element in right.
left and right are non-empty.
left has the smallest possible size.

Return the length of left after such a partitioning. It is guaranteed that such a partitioning exists.

Example 1:

Input: [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]

Example 2:

Input: [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]

Note:

2 <= A.length <= 30000
0 <= A[i] <= 10^6
It is guaranteed there is at least one way to partition A as described.

public int partitionDisjoint(int[] A) {
        int sizeA = A.length;
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        PriorityQueue<Integer> maxHeap=new PriorityQueue<Integer>(sizeA, new Comparator<Integer>() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return o2-o1;
            }
        });

        for (int i = 0; i < sizeA; i++) {
            if (i == 0) maxHeap.add(A[i]);
            else minHeap.add(A[i]);
        }

        // left <= right
        for (;;) {
            int topMinHeap = minHeap.peek();
            int topMaxHeap = maxHeap.peek();
            if (topMaxHeap <= topMinHeap) {
                return maxHeap.size();
            } else {
                int temp = A[maxHeap.size()];
                maxHeap.add(temp);
                minHeap.remove(temp);
            }
        }
    }