标签：level default iss exp for using val pre minutes
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]]
class Solution {
public:
int numberOfBoomerangs(vector<pair<int, int>>& points) {
int size = points.size();
int counter = 0;
for (int i = 0; i < size; ++i) {
for (int j = 0; j < size; ++j) {
if (i == j) continue;
for (int k = 0; k < size; ++k) {
if (i == k || j == k) continue;
int x1 = abs(points[i].first - points[j].first);
int x2 = abs(points[i].first - points[k].first);
int y1 = abs(points[i].second - points[j].second);
int y2 = abs(points[i].second - points[k].second);
double fs = sqrt(pow(x1, 2) + pow(y1, 2));
double ft = sqrt(pow(x2, 2) + pow(y2, 2));
if (fs == ft) counter++;
}
}
}
return counter;
}
};
Approach #2: C++.
class Solution {
public:
int numberOfBoomerangs(vector<pair<int, int>>& points) {
int size = points.size();
int counter = 0;
for (int i = 0; i < size; ++i) {
unordered_map<int, int> distances;
for (int j = 0; j < size; ++j) {
if (i == j) continue;
int distance = (points[i].first - points[j].first) * (points[i].first - points[j].first) + (points[i].second - points[j].second) * (points[i].second - points[j].second);
distances[distance]++;
}
for (auto& p : distances) {
if (p.second > 1)
counter += p.second * (p.second - 1);
}
}
return counter;
}
};
Approach #2: Java.
class Solution {
public int numberOfBoomerangs(int[][] points) {
int res = 0;
for (int i = 0; i < points.length; ++i) {
HashMap<Integer, Integer> map = new HashMap<>();
for (int j = 0; j < points.length; ++j) {
if (points[i] == points[j]) continue;
int dx = points[i][0] - points[j][0];
int dy = points[i][1] - points[j][1];
int d = dx * dx + dy * dy;
map.put(d, map.getOrDefault(d, 0) + 1);
}
for (int val : map.values()) {
res += val * (val - 1);
}
}
return res;
}
}
Approach #3: Python.
class Solution(object):
def numberOfBoomerangs(self, points):
"""
:type points: List[List[int]]
:rtype: int
"""
res = 0
for p in points:
cmap = {}
for q in points:
f = p[0] - q[0]
s = p[1] - q[1]
cmap[f*f + s*s] = 1 + cmap.get(f*f + s*s, 0)
for k in cmap:
res += cmap[k] * (cmap[k] - 1)
return res
| Time Submitted | Status | Runtime | Language |
|---|---|---|---|
| a few seconds ago | Accepted | 241 ms | java |
| 5 minutes ago | Accepted | 1436 ms | python |
| 8 minutes ago | Accepted | 196 ms | cpp |
标签：level default iss exp for using val pre minutes
原文地址：https://www.cnblogs.com/ruruozhenhao/p/9972043.html
