perfect-rectangle

复制代码
https://leetcode.com/problems/perfect-rectangle/

// https://discuss.leetcode.com/topic/55944/o-n-log-n-sweep-line-solution


public class Solution {
    
    public class Column implements Comparable<Column> {
        int xs;
        int[] rect;
    
        public Column(int xs, int[] rect) {
            this.xs = xs;
            this.rect = rect;
        }
    
        public int compareTo(Column that) {
            if (this.xs != that.xs) {
                return this.xs - that.xs;
            }
            return this.rect[0] - that.rect[0];
        }
    
    }

    public boolean isRectangleCover(int[][] rectangles) {
        PriorityQueue<Column> pq = new PriorityQueue<Column>();
        int[] border = {Integer.MAX_VALUE, Integer.MIN_VALUE};
        for (int[] rect : rectangles) {
            Column c1 = new Column(rect[0], rect);
            Column c2 = new Column(rect[2], rect);
            pq.add(c1);
            pq.add(c2);
            if (rect[1] < border[0]) {
                border[0] = rect[1];
            }
            if (rect[3] > border[1]) {
                border[1] = rect[3];
            }
        }
        TreeSet<int[]> tset = new TreeSet<int[]> (new Comparator<int[]>(){
            public int compare(int []rect1, int[]rect2) {
                if (rect1[3] <= rect2[1]) {
                    return -1;
                }
                else if (rect1[1] >= rect2[3]) {
                    return 1;
                }
                else {
                    return 0;
                }
            }
        });
        int yRange = 0;
        while (!pq.isEmpty()) {
            int xs = pq.peek().xs;
            while (!pq.isEmpty() && pq.peek().xs == xs) {
                Column col = pq.poll();
                int[] rect = col.rect;
                if (xs == rect[2]) {
                    tset.remove(rect);
                    yRange -= rect[3] - rect[1];
                }
                else {
                    // xs == rect[0]
                    if (!tset.add(rect)) {
                        // intersect
                        return false;
                    }
                    yRange += rect[3] - rect[1];
                }
            }
            // if pq.isEmpty(), the right line, no need to check
            if (!pq.isEmpty() && yRange != border[1] - border[0]) {
                return false;
            }
        }
        return true;
    }
}
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posted @   blcblc  阅读(185)  评论(0编辑  收藏  举报
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