[leetcode]593. Valid Square

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本文介绍了一种算法，通过计算四个点之间的距离来判断这四个点能否构成一个正方形。该方法利用了正方形边长相等且对角线相等的特点，并通过比较距离的平方来避免开方运算，减少精度损失。

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题目链接：https://leetcode.com/problems/valid-square/#/description

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

思路：

If we calculate all distances between 4 points, 4 smaller distances should be equal (sides), and 2 larger distances should be equal too (diagonals). As an optimization, we can compare squares of the distances, so we do not have to deal with the square root and precision loss.

Therefore, our set will only contain 2 unique distances in case of square (beware of the zero distance though).

class Solution {
public:
    bool validSquare(vector<int>& p1, vector<int>& p2, vector<int>& p3, vector<int>& p4) {
        unordered_set<int> s({d(p1,p2),d(p1,p3),d(p1,p4),d(p2,p3),d(p2,p4),d(p3,p4)});
        return !s.count(0) && s.size()==2;
    }
    int d(vector<int>& p1,vector<int>& p2)
    {
        return pow(p1[0]-p2[0],2)+pow(p1[1]-p2[1],2);
    }
};