LeetCode:447. Number of Boomerangs

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原文链接：http://www.cnblogs.com/luluqiao/p/6363030.html

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#java

本文介绍了一种计算平面上特定回旋镖数量的方法。通过遍历每个点并使用哈希表记录距离出现的次数来找出所有符合要求的回旋镖组合。适用于最多500个坐标点且坐标范围在[-10000,10000]的情况。

package HashTable;
////Question 447. Number of Boomerangs
/*
Given n points in the plane that are all pairwise distinct,
a "boomerang" is a tuple of points (i, j, k) such that the distance between i and j equals
the distance between i and k (the order of the tuple matters).

Find the number of boomerangs.
You may assume that n will be at most 500 and coordinates of points are all
in the range [-10000, 10000] (inclusive).

Example:
Input:
[[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]]
*/

import java.util.HashMap;
import java.util.Map;

public class numberOfBoomerangs447 {
//一开始想的是对称点，后来是直接用最笨的方法，一个一个遍历
public static int numberOfBoomerangs(int[][] points) {
int res=0;
for(int i=0;i<points.length;i++){
Map<Integer,Integer> map=new HashMap<Integer,Integer>();
for(int j=0;j<points.length;j++){
if (i==j)
continue;
int distance=(int) (Math.pow(Math.abs(points[i][0]-points[j][0]),2)+Math.pow(Math.abs(points[i][1]-points[j][1]),2));
map.put(distance,map.getOrDefault(distance,0)+1);
}
for(int val:map.values()){
res+=val*(val-1);
}
map.clear();
}
return res;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int[][] points={{0,1},{0,0},{0,2}};

System.out.println(numberOfBoomerangs(points));
}