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 iand 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:2Explanation:The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

题意：给出n个点的坐标，找到ijk三个点，使得i到j和k的距离相等。

public class Solution {
    public int numberOfBoomerangs(int[][] points) {
        ////还有负数。。。。。hehe....坐标轴。。。。并不是只有x轴
        // int length=points.length;
        // int m;
        // int count=0;
        // for(int i=0;i<length;i++){
        //     m=1;
        //     while(i+m<=length-1 && i-m>=0){
        //         if(points[i][0]-points[i-m][0]==points[i+m][0]-points[i][0]){
        //             count=count+2;
        //         }
        //         m++;
        //     }
        // }
        // // System.out.println(length);
        // return count;
        //http://blog.csdn.net/MebiuW/article/details/53096120
        //  int length=points.length;
        //  int count=0;
        //  for(int i=0;i<length;i++){
        //      HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
        //      for(int j=0;j<length;j++){
        //          int dist=(points[i][0]-points[j][0])*(points[i][0]-points[j][0])+(points[i][1]-points[j][1])*(points[i][1]-points[j][1]);
        //          if(!map.containsKey(dist)){
        //              map.put(dist,0);
        //          }
        //          count+=map.get(dist)*2;
        //          map.put(dist,map.get(dist)+1);
        //      }
        //  }
        //  return count;
        //也是遍历，求与其距离相同的点的个数，最后就是个数（个数-1）
        ///感觉还是这个好理解一些~
        int length=points.length;
        int count=0;
        int dist=0;
        for(int i=0;i<length;i++){
            HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
            for(int j=0;j<length;j++){
                if(i==j){
                    continue;
                }else{
                    dist=(points[i][0]-points[j][0])*(points[i][0]-points[j][0])+(points[i][1]-points[j][1])*(points[i][1]-points[j][1]);
                    if(!map.containsKey(dist)){
                        map.put(dist,1);
                    }else{
                        map.put(dist,map.get(dist)+1);    
                    }
                }
                
            }
        Iterator it = map.keySet().iterator();  
        while(it.hasNext()) {  
            int key = (int)it.next();  
            count+=map.get(key)*(map.get(key)-1);
            // System.out.println("value:" + hashMap.get(key));  
            } 
        }
        return count;
    }
}

PS：第一次以为只有x轴正半轴，然后才发现其实是x,y轴4个空间。

最后的思想是遍历所有点，找到所有点距其距离相等的点的个数n，那么就有n(n-1)个，不断累加即可。

Leetcode 447. Number of Boomerangs JAVA语言