leetcode 593. Valid Square

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: 
<pre style="overflow:auto; font-family:Menlo,Monaco,Consolas,"Courier New",monospace; font-size:13px; padding:9.5px; margin-top:0px; margin-bottom:10px; line-height:1.42857; color:rgb(51,51,51); word-break:break-all; word-wrap:break-word; background-color:rgb(245,245,245); border:1px solid rgb(204,204,204)">Input: p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,1]
Output: True
</pre>
 
 Note:
<ol style="margin-top:0px; margin-bottom:10px; color:rgb(51,51,51); font-family:"Helvetica Neue",Helvetica,Arial,sans-serif; font-size:14px"><li style="">All the input integers are in the range [-10000, 10000].</li><li style="">A valid square has four equal sides with positive length and four equal angles (90-degree angles).</li><li style="">Input points have no order.</li></ol> 就是看4个点是否构成的是正方形。

<pre class="blockcode"><code class="language-java">package leetcode;

import java.util.Arrays;

public class Valid_Square_593 {

public boolean validSquare(int[] p1, int[] p2, int[] p3, int[] p4) {
 int[] allBian=new int[]{
getDistance2(p1, p2),getDistance2(p1, p3),getDistance2(p1, p4),
getDistance2(p2, p3),getDistance2(p2, p4),getDistance2(p3, p4)
 };
 Arrays.sort(allBian);
 if(allBian[0]==0){
return false;
 }
 //保证有4个边相同,a^2可能比a大，也可能比a小
 if(allBian[0]==allBian[3]||allBian[2]==allBian[5]){
//a^2比a大
if(allBian[0]==allBian[3]){
//使用勾股定理:a^2+b^2=c^2，而a=b，所以2a^2=c^2
if(allBian[0]*2==allBian[5]){
 return true;
}
}
else{//a^2比a小
if(allBian[5]*2==allBian[0]){
 return true;
}
}
 }
 return false;
}

//返回距离的平方
int getDistance2(int[] p1,int[] p2){
 int distance2=(p1[0]-p2[0])*(p1[0]-p2[0])+(p1[1]-p2[1])*(p1[1]-p2[1]);
 return distance2;
}

public static void main(String[] args) {
 // TODO Auto-generated method stub
 Valid_Square_593 v = new Valid_Square_593();
 int[] p1 = new int[] { 0, 0 };
 int[] p2 = new int[] { 0, 0 };
 int[] p3 = new int[] { 0, 0 };
 int[] p4 = new int[] { 0, 0 };
 System.out.println(v.validSquare(p1, p2, p3, p4));
}

}</code></pre>我为什么算距离的平方，而不是直接算距离呢？因为一个根号下去，变成double类型，有可能出现精度问题。如根号2=1.4142135623731...，但是1.4142135623731的平方=2.000000004，不等于2。

大神说：如果4点连成的是正方形，那么这4个点之间的6条线，一定有4条相等，另外2条也相等，且4条的长度不等于2条的长度。4条是边，2条是斜边。

<pre class="blockcode"><code class="language-java">public class Solution {
public boolean validSquare(int[] p1, int[] p2, int[] p3, int[] p4) {
 HashMap<Integer, Integer> map = new HashMap<>();
 int a12 = getDistanceSquare(p1, p2);
 int a13 = getDistanceSquare(p1, p3);
 int a14 = getDistanceSquare(p1, p4);
 int a23 = getDistanceSquare(p2, p3);
 int a24 = getDistanceSquare(p2, p4);
 int a34 = getDistanceSquare(p3, p4);
 if (a12 == 0 || a13 == 0 || a14 == 0 || a23 == 0 || a24 == 0 || a34 == 0){
 return false;
 }
 map.put(a12, map.getOrDefault(a12, 0) + 1);
 map.put(a13, map.getOrDefault(a13, 0) + 1);
 map.put(a14, map.getOrDefault(a14, 0) + 1);
 map.put(a23, map.getOrDefault(a23, 0) + 1);
 map.put(a24, map.getOrDefault(a24, 0) + 1);
 map.put(a34, map.getOrDefault(a34, 0) + 1);
 return map.size() == 2 && (map.get(a12) == 2 || map.get(a12) == 4);

}

private int getDistanceSquare(int[] a, int[] b) {
 return (a[0] - b[0]) * (a[0] - b[0]) + (a[1] - b[1]) * (a[1] - b[1]);
}
}</code></pre>这道题有solutions：
<a href="https://leetcode.com/problems/valid-square/solution/" rel="noopener noreferrer" target="_blank">https://leetcode.com/problems/valid-square/solution/</a>。

<h2 id="solution" style="font-family:"Helvetica Neue",Helvetica,Arial,sans-serif; line-height:1.7; color:rgb(51,51,51); ma