?LeetCode刷題實(shí)戰(zhàn)391：完美矩形

示例 1:
rectangles = [
  [1,1,3,3],
  [3,1,4,2],
  [3,2,4,4],
  [1,3,2,4],
  [2,3,3,4]
]
返回 true。5個(gè)矩形一起可以精確地覆蓋一個(gè)矩形區(qū)域。

示例 2:
rectangles = [
  [1,1,2,3],
  [1,3,2,4],
  [3,1,4,2],
  [3,2,4,4]
]
返回 false。兩個(gè)矩形之間有間隔，無法覆蓋成一個(gè)矩形。

示例 3:
rectangles = [
  [1,1,3,3],
  [3,1,4,2],
  [1,3,2,4],
  [3,2,4,4]
]
返回 false。圖形頂端留有間隔，無法覆蓋成一個(gè)矩形。

示例 4:
rectangles = [
  [1,1,3,3],
  [3,1,4,2],
  [1,3,2,4],
  [2,2,4,4]
]

返回 false。因?yàn)橹虚g有相交區(qū)域，雖然形成了矩形，但不是精確覆蓋。

class Solution {
    public boolean isRectangleCover(int[][] rectangles) {
        // 完美矩形的左下角和右上角坐標(biāo)
        int X1 = Integer.MAX_VALUE, Y1 = Integer.MAX_VALUE;
        int X2 = Integer.MIN_VALUE, Y2 = Integer.MIN_VALUE;
    
        // 小矩形面積之和
        int areas = 0;
        // 記錄所有頂點(diǎn)的出現(xiàn)情況
        Set<String> points = new HashSet<>();
        for (int[] rectangle : rectangles) {
            int x1 = rectangle[0], y1 = rectangle[1], x2 = rectangle[2], y2 = rectangle[3];
            // 更新坐標(biāo)
            X1 = Math.min(x1, X1);
            Y1 = Math.min(y1, Y1);
            X2 = Math.max(x2, X2);
            Y2 = Math.max(y2, Y2);

            areas += (x2 - x1) * (y2 - y1);
            // 判斷頂點(diǎn)是否出現(xiàn)過
            String[] ps = {x1 + " " + y1, x2 + " " + y2, x1 + " " + y2, x2 + " " + y1};
            for (String s : ps) {
                // 沒有出現(xiàn)過，添加；否則，移除
                if(points.contains(s)){
                    points.remove(s);
                } else {
                    points.add(s);
                }
            }
        }
    
        // 面積是否相等 
        int expected = (X2 - X1) * (Y2 -Y1);
        if(areas != expected){
            return false;
        }
        // 頂點(diǎn)情況是否滿足
        if(points.size() != 4 || !points.contains(X1 + " " + Y1) || !points.contains(X2 + " " + Y2) || !points.contains(X1 + " " + Y2) || !points.contains(X2 + " " + Y1)){
            return false;
        }
        return true;
    }
}