- 题目链接850. Rectangle Area II
- 题目意思
- 给出一些矩形,这些矩形可能会重叠,求矩形所围成图形的面积。
- 题目解答
- 步骤
- 延y轴扫描
- y = 0, 底 3 = 3 - 0, 高 y = 0,
- y = 1, 底 2 = 2 - 0, 高 y = 1, 面积 前底 3 x (当前高1 - 前高0) = 3;
- y = 2, 底 2 = 2 - 0, 高 y = 2, 面积 前底 2 x (当前高2 - 前高1) = 2;
- …
- 怎么算底呢?我们可以用合并区间的方式,如合并y=0时的各个底边(0,2), (1,2), (1,3),可以按照最前的元素排序区间,在合并这些区间。
public int rectangleArea(int[][] rectangles) {
int OPEN = 0, CLOSE = 1;
int[][] events = new int[rectangles.length * 2][];
int t = 0;
for (int[] rect : rectangles) {
events[t++] = new int[]{rect[1], OPEN, rect[0], rect[2]};
events[t++] = new int[]{rect[3], CLOSE, rect[0], rect[2]};
}
Arrays.sort(events, (a,b)->Integer.compare(a[0], b[0]));
List<int[]> active = new ArrayList<>();
int cur_y = events[0][0];
long res = 0;
for (int[] event : events) {
int y = event[0], type = event[1], x1 = event[2], x2 = event[3];
long query = 0;
int cur = -1;
for (int[] act : active) {
cur = Math.max(cur, act[0]);
query += Math.max(act[1] - cur, 0);
cur = Math.max(cur, act[1]);
}
res += query * (y - cur_y);
if (type == OPEN) {
active.add(new int[]{x1, x2});
Collections.sort(active, (a, b) -> Integer.compare(a[0], b[0]));
} else {
for (int i = 0; i < active.size(); i++) {
if (active.get(i)[0] == x1 && active.get(i)[1] == x2) {
active.remove(i);
break;
}
}
}
cur_y = y;
}
res %= 1_000_000_007;
return (int)res;
}