leetcode939. Minimum Area Rectangle

Posted 2021-01-29 seyjs

题目如下：

Given a set of points in the xy-plane, determine the minimum area of a rectangle formed from these points, with sides parallel to the x and y axes.

If there isn‘t any rectangle, return 0.

Example 1:

Input: [[1,1],[1,3],[3,1],[3,3],[2,2]]
Output: 4

Example 2:

Input: [[1,1],[1,3],[3,1],[3,3],[4,1],[4,3]]
Output: 2

Note:

1. 1 <= points.length <= 500
2. 0 <= points[i][0] <= 40000
3. 0 <= points[i][1] <= 40000
4. All points are distinct.

解题思路：题目中约定了矩形的各边要与X轴或者Y轴平行，所以不妨把所有的点按X轴进行分组。例如 [[1,1],[1,3],[3,1],[3,3],[2,2]]，以X的坐标分组后得到 {1: [1, 3], 2: [2], 3: [1, 3]}，接下来只要判断两个不同的X轴所对应的Y轴的list中是否存在相同的元素，如果相同则表示可以组成一个题目中要求的矩形。最后求出最小面积即可。

代码如下：

class Solution(object):
    def minAreaRect(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        dic_x = {}
        res = float(‘inf‘)
        x_set = set()
        for (x,y) in points:
            dic_x[x] = dic_x.setdefault(x,[]) + [y]
            x_set.add(x)
        x_list = sorted(list(x_set))

        for kx1 in range(len(x_list)):
            for kx2 in range(kx1+1,len(x_list)):
                intersection = sorted((list(set(dic_x[x_list[kx1]]) & set(dic_x[x_list[kx2]]))))
                if len(intersection) < 2:
                    continue
                minDis = intersection[-1] - intersection[0]
                for i in range(len(intersection)-1):
                    minDis = min(minDis,intersection[i+1] - intersection[i])
                #print kx1,kx2,intersection
                res = min(res, minDis * abs(x_list[kx1]-x_list[kx2]))
        return res if res != float(‘inf‘) else 0