[LeetCode 447.] Number of Boomerangs
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LeetCode 447. Number of Boomerangs
题目描述
You are given n points in the plane that are all distinct, where points[i] = [xi, yi]. 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).
Return the number of boomerangs.
Example 1:
Input: points = [[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]].
Example 2:
Input: points = [[1,1],[2,2],[3,3]]
Output: 2
Example 3:
Input: points = [[1,1]]
Output: 0
Constraints:
- n == points.length
- 1 <= n <= 500
- points[i].length == 2
- -104 <= xi, yi <= 104
- All the points are unique.
解题思路
题目说的是从平面上不重复的一组点中,任意选取三个点可以组成一个回旋镖,问能选出几组。注意,这里的回旋镖把同一个回旋镖计算了两次。
这里如果直接三重循环遍历,会超时。解决办法是利用哈希表降低复杂度。这里我们直接枚举回旋镖的中点,然后遍历其两个尾翼即可。
参考代码
/*
* @lc app=leetcode id=447 lang=cpp
*
* [447] Number of Boomerangs
*/
// @lc code=start
class Solution {
public:
int numberOfBoomerangs(vector<vector<int>>& points) {
int res = 0;
unordered_map<int, int> hsmap;
for (auto&& p : points) {
hsmap.clear();
for (auto&& q : points) {
int dist2 = (p[0]-q[0])*(p[0]-q[0]) + (p[1]-q[1])*(p[1]-q[1]);
hsmap[dist2]++;
}
for (auto&& [k, v]: hsmap) {
res += v*(v-1);
}
} // 哈希表把枚举时间复杂度从 O(n^3) 降到 O(n^2)
return res;
} // AC
};
// @lc code=end
标签:lc,Boomerangs,447,Example,points,auto,hsmap,LeetCode 来源: https://www.cnblogs.com/zhcpku/p/15261945.html