963. Minimum Area Rectangle II

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Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes.

If there isn't any rectangle, return 0.

Example 1:

Input: [[1,2],[2,1],[1,0],[0,1]]
Output: 2.00000
Explanation: The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2.

Example 2:

Input: [[0,1],[2,1],[1,1],[1,0],[2,0]]
Output: 1.00000
Explanation: The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1.

Example 3:

Input: [[0,3],[1,2],[3,1],[1,3],[2,1]]
Output: 0
Explanation: There is no possible rectangle to form from these points.

Example 4:

Input: [[3,1],[1,1],[0,1],[2,1],[3,3],[3,2],[0,2],[2,3]]
Output: 2.00000
Explanation: The minimum area rectangle occurs at [2,1],[2,3],[3,3],[3,1], with an area of 2.

Note:

1. 1 <= points.length <= 50
2. 0 <= points[i][0] <= 40000
3. 0 <= points[i][1] <= 40000
4. All points are distinct.
5. Answers within 10^-5 of the actual value will be accepted as correct.

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思路：暴力求解，遍历矩形3个点的位置，另外一个点可以通过已知的3个点求出来，注意：

（1）水平线，竖直线斜率的求解

（2）Python做些prune才能AC，可以Run Code看超时case的运行时间

import math
class Solution(object):
    def minAreaFreeRect(self, points):
        """
        :type points: List[List[int]]
        :rtype: float
        """
        s = set([tuple(t) for t in points])
        res = float('inf')
        n=len(points)
        for i in range(n):
            xi,yi = points[i]
            for j in range(i+1, n):
#                if j==i: continue
                xj,yj = points[j]
                if (xj-xi)**2+(yj-yi)**2>=res: continue
                k_ij = (yj-yi)/(xj-xi+0.0) if xi!=xj else float('inf')
                for k in range(n):
                    if k==i or k==j: continue
                    xk,yk = points[k]
                    k_jk = (yk-yj)/(xk-xj+0.0) if xk!=xj else float('inf')
                    if (k_ij,k_jk)==(float('inf'),0) or (k_ij,k_jk)==(0,float('inf')):
                        pass
                    else:
                        if abs(-1-k_ij*k_jk)>0.00001: continue
                    
                    xl,yl=(xi+xk-xj),(yi+yk-yj)
                    if (xl,yl) in s:
                        a2 = (xj-xi)**2+(yj-yi)**2
                        b2 = (xk-xj)**2+(yk-yj)**2
                        res = min(res, a2*b2)
        return math.sqrt(res) if res!=float('inf') else 0