Leetcode算法学习日志-447 Number of Boomerangs
2017-10-30 15:10
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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
iand
jequals the distance between
iand
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]]
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)。
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