LeetCode 218: The Skyline Problem

最新推荐文章于 2021-11-21 01:10:42 发布

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最新推荐文章于 2021-11-21 01:10:42 发布

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原文链接：http://www.cnblogs.com/shuashuashua/p/7473345.html

本文介绍了一种计算由多个矩形建筑物组成的天际线轮廓的高效算法。通过记录每个建筑物的高度变化并使用优先队列来跟踪当前最高的建筑物，该算法能够正确地生成天际线轮廓的关键转折点。

Note:

1. There is a trick here: As it only gets the starting point, we can check each bound of the building. Also we have to have a "sign" to know when the bound is start when the bound is end.

2. REMEMBER to add 0 to the queue to avoid NPE. Also 0 represents there is no building there.

class Solution {
    public List<int[]> getSkyline(int[][] buildings) {
        if (buildings.length == 0) {
            return new ArrayList<>();
        }
        
        List<int[]> result = new ArrayList<>();
        List<int[]> height = new ArrayList<>();
        for (int[] building : buildings) {
            height.add(new int[]{building[0], -building[2]});
            height.add(new int[]{building[1], building[2]});
        }
        
        Collections.sort(height, (b1, b2) -> {return b1[0] == b2[0] ? b1[1] - b2[1] : b1[0] - b2[0];});
        PriorityQueue<Integer> queue = new PriorityQueue<>((i1, i2) -> (i2 - i1));
        queue.offer(0);
        int prev = 0;
        for (int[] vertical : height) {
            if (vertical[1] < 0) {
                queue.offer(-vertical[1]);
            } else {
                queue.remove(vertical[1]);
            }
            
            int current = queue.peek();
            if (prev != current) {
                result.add(new int[]{vertical[0], current});
                prev = current;
            }
        }
        return result;
    }
}