447. Number of Boomerangs
題目連結:https://leetcode.com/problems...
遍歷每個point,然後找和它等距離的其他point,按距離來儲存,比如有一個點a,和它距離都是1的點有b,c,d,那麼一共的組合就有6種,包括:[a, b, c], [a, c, b], [a, b, d], [a, d, b], [a, c, d], [a, d, c]。這麼算是不考重複的情況下。還有可能b, c座標完全相同,那麼b和c會被當成兩個點算兩次。
public class Solution {
public int numberOfBoomerangs(int[][] points) {
// traverse i, find the distance, keep the same distance in hashmap
int result = 0;
for(int i = 0; i < points.length; i++) {
Map<Integer, Integer> map = new HashMap();
for(int j = 0; j < points.length; j++) {
if(i == j) continue;
int dx = points[j][0] - points[i][0], dy = points[j][1] - points[i][1];
int distance = dx * dx + dy * dy;
map.put(distance, map.getOrDefault(distance, 0) + 1);
}
for(int k : map.keySet()) {
int n = map.get(k);
result += n * (n - 1);
}
}
return result;
}
}