447. Number of Boomerangs

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]]

Solution1:(TLE)

class Solution:
    def numberOfBoomerangs(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        res = 0
        def compute(x,y):
            return (x[0]-y[0])**2 + (x[1]-y[1])**2
        for i in range(len(points)-2):
            for j in range(i+1,len(points)-1):
                for k in range(j+1,len(points)):
                    if compute(points[i],points[j]) == compute(points[i],points[k]):
                        res += 2
                    if compute(points[i],points[j]) == compute(points[j],points[k]):
                        res += 2
                    if compute(points[k],points[j]) == compute(points[i],points[k]):
                        res += 2
        return res

Brute force.
24 / 31 test cases passed.

Solution2:

class Solution:
    def numberOfBoomerangs(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        res = 0
        def compute(x,y):
            return (x[0]-y[0])**2 + (x[1]-y[1])**2
        for i in range(len(points)):
            dic = {}
            for j in range(len(points)):
                if i==j:
                    continue
                temp = compute(points[i],points[j])
                if temp not in dic:
                    dic[temp] = 1
                else:
                    dic[temp] += 1
            for key,value in dic.items():
                res += value*(value-1)
        return res

典型的以空间换时间的策略，用一个字典保存每个点到该点的距离，从而将时间复杂度从O(n3)降到了O(n2)。