标签：continue nat csharp public fbo sum pre ane mos
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:
Explanation:
The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]
public int numberOfBoomerangs(int[][] points) {
int res = 0;
Map<Integer, Integer> map = new HashMap<>();
for(int i=0; i<points.length; i++) {
for(int j=0; j<points.length; j++) {
if(i == j)
continue;
int d = getDistance(points[i], points[j]);
map.put(d, map.getOrDefault(d, 0) + 1);
}
for(int val : map.values()) {
res += val * (val-1);
}
map.clear();
}
return res;
}
private int getDistance(int[] a, int[] b) {
int dx = a[0] - b[0];
int dy = a[1] - b[1];
return dx*dx + dy*dy;
}
标签：continue nat csharp public fbo sum pre ane mos
原文地址：http://www.cnblogs.com/apanda009/p/7789107.html
