The effective square of the test (one question of the day 7/29)

I am using complete computational brute force for this question.

The topic address is as follows
链接

题目要求如下

给定2D空间中四个点的坐标 p1, p2, p3 和 p4,如果这四个点构成一个正方形,则返回 true .
点的坐标 pi 表示为 [xi, yi] .输入 不是 按任何顺序给出的.
一个 有效的正方形 有四条等边和四个等角(90度角).
来源：力扣（LeetCode）

这是基本的要求.其实呢！Prompt I did not use,Because I'm doing purely mathematical calculations,The feature of coordinates is used.比较暴力,The code is fast,But the efficiency is high.

下面是我的代码

class Solution {
    
    public boolean validSquare(int[] p1, int[] p2, int[] p3, int[] p4) {
    
           boolean result = judge_finally(p1,p2,p3,p4);
           return result;

    }
    public static boolean judge_finally(int[]p1,int[]p2,int[]p3,int[]p4) {
    
        double v_1_2 = judge_rhomb(p1, p2);//Some possible distances are calculated here
        double v2_3 = judge_rhomb(p2, p3);
        double v3_4 = judge_rhomb(p3, p4);
        double v1_4 = judge_rhomb(p1, p4);
        double v_2_4 = judge_rhomb(p2, p4);
        double v_1_3 = judge_rhomb(p1, p3);
        
        //Here is a list of possible diagonals,And make a diagonal judgment,Including the coincidence of the four-point diagonal midpoints,Confirm the quad
        //Equal distance and diagonal vertical constraints.This progressively constrains it to a square
        //Note the constraint zero coordinates,So a constraint must be made,The diagonal distance is constrained here0

        if (v_1_3 == v_2_4 && v_1_3 != 0&&judge_vertical(p1, p3, p2, p4)&judge_if_line(p1,p3,p2,p4)) {
    
              return true;
        }
        if (v1_4 == v2_3 && v1_4 != 0&&judge_vertical(p1, p4, p2, p3)&judge_if_line(p1,p4,p2,p3)) {
    
            return true;
        }
        if (v_1_2 == v3_4 && v_1_2 != 0&&judge_vertical(p1, p2, p3, p4)&judge_if_line(p1,p2,p3,p4)) {
    
            return true;
        }
        return  false;

    }

    public static double judge_rhomb(int[]a,int[]b)//Calculate the length of the line formed by the two coordinates
    {
    
       double x1 = ( Math.pow(a[0]-b[0],2)+Math.pow(a[1]-b[1],2));
       return  x1;
    }

    public static boolean judge_vertical(int[]a,int[]b,int[]c,int[]d)//判断直角
    {
    
        int v[] = {
    a[0]-b[0],a[1]-b[1]};
        int v1[] = {
    c[0]-d[0],c[1]-d[1]};
        return  v[0]*v1[0]+v[1]*v1[1]==0;

    }
    public static boolean judge_if_line(int[]a,int[]b,int[]c,int[]d)//Determine whether four points can form a parallelogram
        //If four points can form a parallelogram,The diagonal midpoints coincide
        
    {
    
        boolean bool_line_ = a[0] + b[0] == c[0] + d[0];
        boolean bool_line__ = a[1]+b[1]==c[1]+d[1];
        if (bool_line_&&bool_line__)
        {
    
            return true;
        }
        return false;

    }
}

Reduction method,很基础的方法
调用判断.The vertical here uses the characteristics of the vector.
x1x2+y1y2=0,In this way, you can determine whether it is vertical or not.
Judging whether the four coordinate points constitute a parallelogram,It is necessary to judge whether the midpoints of the diagonals of the possible cases coincide,Then it is only necessary to calculate whether the horizontal and vertical coordinates are equal to each other.
The diagonals of a rhombus bisect each other perpendicularly,We just have to make the diagonals equal,It can be judged as a square.But be careful to do these given coordinate constraints.So I gave the diagonal length not for0.Because if the horizontal and vertical coordinates of the four points are 0的话,The above requirements are actually met.

Complete diagonal constraints.