LeetCode 力扣每日一题 题解 391. 完美矩形

原题链接：391. 完美矩形

题目描述：

给你一个数组 rectangles ，其中 rectangles[i] = [xi, yi, ai, bi] 表示一个坐标轴平行的矩形。这个矩形的左下顶点是 (xi, yi) ，右上顶点是 (ai, bi) 。

如果所有矩形一起精确覆盖了某个矩形区域，则返回 true ；否则，返回 false 。（难度：困难）

C++代码：

typedef pair Point;

class Solution {
public:
    bool isRectangleCover(vector>& rectangles) {
        long area = 0;
        int minX = rectangles[0][0], minY = rectangles[0][1], maxX = rectangles[0][2], maxY = rectangles[0][3];
        map cnt;
        for (auto & rect : rectangles) {
            int x = rect[0], y = rect[1], a = rect[2], b = rect[3];
            area += (long) (a - x) * (b - y);

            minX = min(minX, x);
            minY = min(minY, y);
            maxX = max(maxX, a);
            maxY = max(maxY, b);

            Point point1({x, y});
            Point point2({x, b});
            Point point3({a, y});
            Point point4({a, b});

            cnt[point1] += 1;
            cnt[point2] += 1;
            cnt[point3] += 1;
            cnt[point4] += 1;
        }

        Point pointMinMin({minX, minY});
        Point pointMinMax({minX, maxY});
        Point pointMaxMin({maxX, minY});
        Point pointMaxMax({maxX, maxY});
        if (area != (long long) (maxX - minX) * (maxY - minY) || !cnt.count(pointMinMin) || !cnt.count(pointMinMax) || !cnt.count(pointMaxMin) || !cnt.count(pointMaxMax)) {
            return false;
        }

        cnt.erase(pointMinMin);
        cnt.erase(pointMinMax);
        cnt.erase(pointMaxMin);
        cnt.erase(pointMaxMax);

        for (auto & entry : cnt) {
            int value = entry.second;
            if (value != 2 && value != 4) {
                return false;
            }
        }
        return true;
    }
};