难度: Medium
原题连接
内容描述
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 <= points.length <= 50
0 <= points[i][0] <= 40000
0 <= points[i][1] <= 40000
All points are distinct.
Answers within 10^-5 of the actual value will be accepted as correct.
思路 1 - 时间复杂度: O(N^3)- 空间复杂度: O(N)******
现将points全部存到lookup里面
三轮循环，每轮循环分别取一个点，p1,p2,p3，然后看看直线p1-p2和直线p1-p3是否是互相垂直的，因为我们打算让p2和p3作为对角线 （其实一共有三种组合方式，p2-p3,p1-p2,p1-p3都可以作为对角线）
- 如果互相垂直，那么算出最后一个点的坐标看看是否在lookup里面
- 如果在，算出面积，更新
- 如果不在，continue
- 如果不相互垂直，continue
class Solution(object):
def minAreaFreeRect(self, points):
"""
:type points: List[List[int]]
:rtype: float
"""
lookup = {}
for p in points:
x, y = p
lookup[(x,y)] = 1
res = sys.maxsize
for i in range(len(points)):
x1, y1 = points[i]
for j in range(i+1, len(points)):
x2, y2 = points[j]
for k in range(j+1, len(points)):
x3, y3 = points[k]
if self.isValidRectangle(x1, x2, x3, y1, y2, y3):
x4, y4 = self.getP4(x1, x2, x3, y1, y2, y3)
if (x4, y4) in lookup:
res = min(res, self.getArea(x1, x2, x3, y1, y2, y3))
if self.isValidRectangle(x2, x1, x3, y2, y1, y3):
x4, y4 = self.getP4(x2, x1, x3, y2, y1, y3)
if (x4, y4) in lookup:
res = min(res, self.getArea(x2, x1, x3, y2, y1, y3))
if self.isValidRectangle(x3, x2, x1, y3, y2, y1):
x4, y4 = self.getP4(x3, x2, x1, y3, y2, y1)
if (x4, y4) in lookup:
res = min(res, self.getArea(x3, x2, x1, y3, y2, y1))
return res if res != sys.maxsize else 0
# line p1-p2 and line p1-p3 are Perpendicular to each other
def isValidRectangle(self, x1, x2, x3, y1, y2, y3):
return (x2-x1) * (x3-x1) + (y2-y1) * (y3-y1) == 0
def getP4(self, x1, x2, x3, y1, y2, y3):
return [x3 - x1 + x2, y3 - y1 + y2]
def getArea(self, x1, x2, x3, y1, y2, y3):
return math.sqrt((x2-x1)**2 + (y2-y1)**2) * math.sqrt((x3-x1)**2 + (y3-y1)**2)思路 2 - 时间复杂度: O(N^2)- 空间复杂度: O(N)******
其实这个时间可以得到优化，我们可以做两次循环，每次取得一个点，p1,p2，我们认为p1和p2组成的连线就是对角线
我们将一条对角线的距离和其中点的坐标作为key，p1和p2在points中的index作为value存起来，value默认是一个list
这样如果有一个key的value的长度大于1，那么就说明有两条不同的直线，他们的长度相等并且连线的中点也一样，那么这两条对角线所对应的4个点肯定可以组成一个矩形
那么我们拿出一条对角线，取得上面的两个点p1,p2，然后从另外一条对角线上取得一个点p3,直线p1-p2是对角线，p3到p1和p2的距离即为矩形的长和宽, 求出面积之后更新res即可
这样我们的时间就得到了优化，O(N^2)
class Solution(object):
def minAreaFreeRect(self, points):
"""
:type points: List[List[int]]
:rtype: float
"""
lookup = collections.defaultdict(list)
for i in range(len(points)):
x1, y1 = points[i]
for j in range(i+1, len(points)):
x2, y2 = points[j]
dist = (x1-x2) ** 2 + (y1-y2) ** 2
cx, cy = (x1 + x2) / 2, (y1 + y2) / 2
key = (dist, cx, cy)
lookup[key].append((i, j))
res = sys.maxsize
for k, v in lookup.items():
if len(v) > 1:
for i in range(len(v)):
for j in range(i+1, len(v)):
x1, y1 = points[v[i][0]]
x2, y2 = points[v[i][1]]
x3, y3 = points[v[j][0]] # 直线p1-p2是对角线，p3到p1和p2的距离即为矩形的长和宽
length = math.sqrt((x3-x1)**2 + (y3-y1)**2)
width = math.sqrt((x3-x2)**2 + (y3-y2)**2)
area = length * width
res = min(res, area)
return res if res != sys.maxsize else 0