原題:
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]]
題意:要求計算所有滿足到點i的距離相等的有序元組[i, j, k],其中i、j、k代表座標點。
解決方案:對每個點i,計算其他所有點到點i的距離,並用一個map容器m存儲每個距離所對應的點的個數,對於每個距離d,每新增一個點j使得i與j之間的距離等於d,只需將ans加上2*m[d](由於j、k有序),並將m[d]++即可。
代碼實現如下:
class Solution {
public:
int numberOfBoomerangs(vector<pair<int, int> >& points) {
int n = points.size(), ans = 0;
for(int i=0; i<n; i++)
{
map<int, int> m;
for(int j=0; j<n; j++)
{
int dx = points[i].first-points[j].first, dy = points[i].second-points[j].second;
ans += m[dx*dx + dy*dy] ++;
}
}
return 2*ans;
}
};