number-of-boomerangs

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cangjie

Algorithms

You are given <code>n</code> <code>points</code> in the plane that are all distinct, where <code>points[i] = [xi, yi]</code>. A boomerang is a tuple of points <code>(i, j, k)</code> such that the distance between <code>i</code> and <code>j</code> equals the distance between <code>i</code> and <code>k</code> (the order of the tuple matters).

Return the number of boomerangs.

&nbsp;
Example 1:

<pre>
Input: points = [[0,0],[1,0],[2,0]]
Output: 2
Explanation: The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]].
</pre>

Example 2:

<pre>
Input: points = [[1,1],[2,2],[3,3]]
Output: 2
</pre>

Example 3:

<pre>
Input: points = [[1,1]]
Output: 0
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>n == points.length</code></li>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>points[i].length == 2</code></li>
	<li><code>-104 &lt;= xi, yi &lt;= 104</code></li>
	<li>All the points are unique.</li>
</ul>

numberOfBoomerangs

Number of Boomerangs

Line Reflection

line-reflection

直线镜像

Array

Hash Table

Math

给定平面上&nbsp;<code>n</code> 对 互不相同 的点&nbsp;<code>points</code> ，其中 <code>points[i] = [xi, yi]</code> 。回旋镖 是由点&nbsp;<code>(i, j, k)</code> 表示的元组 ，其中&nbsp;<code>i</code>&nbsp;和&nbsp;<code>j</code>&nbsp;之间的欧式距离和&nbsp;<code>i</code>&nbsp;和&nbsp;<code>k</code>&nbsp;之间的欧式距离相等（需要考虑元组的顺序）。

返回平面上所有回旋镖的数量。
&nbsp;

示例 1：

<pre>
输入：points = [[0,0],[1,0],[2,0]]
输出：2
解释：两个回旋镖为 [[1,0],[0,0],[2,0]] 和 [[1,0],[2,0],[0,0]]
</pre>

示例 2：

<pre>
输入：points = [[1,1],[2,2],[3,3]]
输出：2
</pre>

示例 3：

<pre>
输入：points = [[1,1]]
输出：0
</pre>

&nbsp;

提示：

<ul>
	<li><code>n ==&nbsp;points.length</code></li>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>points[i].length == 2</code></li>
	<li><code>-104 &lt;= xi, yi &lt;= 104</code></li>
	<li>所有点都 互不相同</li>
</ul>