Leetcode算法学习日志-447 Number of Boomerangs

Leetcode 447 Number of Boomerangs

题目原文

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 betweeni and k (the order of the tuple matters).

Find the number of boomerangs. You may assume thatn 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:2Explanation:The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

题意分析

给定n个点及它们的坐标，求出"boomerangs"的个数，（i，j，k）一组点，如果i到j的距离和i到k的距离一样，则这样一组点称为"boomerangs"，j和k的顺序可以调换得到另一个“boomerangs”。

解法分析

要从n个点中选取3个点，不同的选取方法（考虑顺序）个数为O(n^3)，因此算法能达到O(n^2)即可。对于每一个点，遍历其他点，将距离作为关键字，有相应距离的点的个数为关联值，建立一个unordered_map，对于本题的情况，unordered_map采用直接寻址表，查询、插入、搜索操作的复杂度都是O(1)，就算是桶内有多个元素的采用链接法的散列表，如果哈希函数选得均匀，字典操作复杂度也为O(1)。建好表后，用范围for语句遍历表，对于大于一的关联值求增加的“boomerangs”数。C++代码如下：

class Solution {public:    int numberOfBoomerangs(vector<pair<int, int>>& points) {        int res=0;        unordered_map<int,int> distance;//first is the distance, second is the number of points that the same distance with point i        int i,j,key;        for(i=0;i<points.size();i++){            distance.clear();            for(j=0;j<points.size();j++)            {                if(j==i)                    continue;                key=(points[i].first-points[j].first)*(points[i].first-points[j].first)+(points[i].second-points[j].second)*(points[i].second-points[j].second);                distance[key]++;            }            for(auto p:distance){                if(p.second>1)                    res+=p.second*(p.second-1);            }        }        return res;            }};

由于map的操作复杂度都是O(1)，因此算法复杂度为O(n^2)。