<?php
/*功能:求随意四个点是否能组成四边形给你四个坐标点,判断它们能不能组成一个矩形,如判断([0,0],[0,1],[1,1],[1,0])能组成一个矩形。
我们分析这道题, 给4个标点,判断是否矩形
高中知识,矩形有4条边,两两相等, 矩形两条对角线相等, 矩形的长短边与对角线满足勾股定理。
故解题思路为,根据坐标点,
列出所有的两点组合边长的数组,去重,看是不是只剩 3个长度(注意正方形2个长度)
判断是否满足勾股定理
调优一下,先判断有没有重复的点,有的话肯定不是矩形
*/
代码如下:
function isRectangle($point1, $point2, $point3, $point4){
if ($point1 == $point2 || $point1 == $point3 || $point1 == $point4 || $point2 == $point3 || $point2 == $point4 || $point3 == $point4) { return false; } $lengthArr = []; $lengthArr[] = getLengthSquare($point1, $point2); $lengthArr[] = getLengthSquare($point1, $point3); $lengthArr[] = getLengthSquare($point1, $point4); $lengthArr[] = getLengthSquare($point2, $point3); $lengthArr[] = getLengthSquare($point2, $point4); $lengthArr[] = getLengthSquare($point3, $point4); $lengthArr = array_unique($lengthArr); $lengthCount = count($lengthArr); if ($lengthCount == 3 || $lengthCount == 2 ) { if ($lengthCount == 2) { return(max($lengthArr) == 2*min($lengthArr)); } else { $maxLength = max($lengthArr); $minLength = min($lengthArr); $otherLength = array_diff($lengthArr, [$maxLength, $minLength]); return($minLength + $otherLength == $maxLength); } } else { return false; } } function getLengthSquare($point1, $point2){ $res = pow($point1[0]-$point2[0], 2)+pow($point1[1]-$point2[1], 2); return $res; } var_dump(isRectangle([0,0],[0,1],[1,1],[1,0]));感谢https://www.cnblogs.com/jwcrxs/p/8986120.html此博客的分享