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進入正題

Given the coordinates of four points in 2D space, return whether the four points could construct a square.
The coordinate (x,y) of a point is represented by an integer array with two integers.
Example:
Input: p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,1]
Output: True
Note:
All the input integers are in the range [-10000, 10000].
A valid square has four equal sides with positive length and four equal angles (90-degree angles).
Input points have no order.

我想到的方法是，把p1想成固定的起始點，一個個和另外三個點比，找看有沒有x座標或y座標有相同的作為下個點；
找到第二個點之後，再和另外兩個點相比確認，依序確認成正方形
但是這樣程式碼的判斷會很多很冗長，而且無法判斷出兩邊不是平行於X軸和Y軸的正方形

正好跟我爸討論了下，想到可以檢查任兩點的距離是否為A或根號2倍A
這樣程式碼應該像這樣:

def validSquare(self, p1: List[int], p2: List[int], p3: List[int], p4: List[int]) -> bool:
    Ip1p2 = distance(p1,p2) 
    Jp1p3 = distance(p1,p3)
    Kp1p4 = distance(p1,p4)
    if (Ip1p2 == Jp1p3 and Ip1p2 == pow(Kp1p4, 0.5)) or (Ip1p2 == Jp1p3 and Jp1p3 == pow(Kp1p4, 0.5)) or (Jp1p2 == Kp1p3 and Jp1p3 == pow(Ip1p4, 0.5)):
        return True
    else:
        return False
    
def distance(self, A:List[int], B:List[int]) -> float:
    dist = pow((pow((A[0]-B[0]), 2) + pow((A[1]-B[1]), 2), 0.5)
    if ((A[0]-B[0])<0) or ((A[1]-B[1])<0):
        return (0 - dist)
    else:
        return dist

結果在這行
if ((A[0]-B[0])<0) or ((A[1]-B[1])<0):
出現了語法錯誤，明天繼續

參考資料