447. Number of boomerangs Boomerang array

Source: Internet
Author: User

[Copy question]:

Given n points in the plane that was all pairwise distinct, a "boomerang" was a tuple of points (i, j, k) such that th e distance between i and j equals the distance between and (the order of the i k tuple matters).

Find the number of boomerangs. Assume that n would be is at most and coordinates of points is all in the range [ -10000, 10000] (Inclusi VE).

Example:

Input:[[0,0],[1,0],[2,0]]output:2explanation:the boomerangs is [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

[Brute force solution]:

Time Analysis:

Spatial Analysis:

[After optimization]:

Time Analysis: n^2

Spatial Analysis:

[Wonderful output CONDITIONS]:

[Wonderful corner case]:

[Thinking questions]:

Do not know how to indicate the distance is equal: The number of occurrences of the statistical distance with HashMap

[a sentence of thought]:

Using HashMap to calculate the number of occurrences of distance, then select 2 points with the combination formula

[input]: null: Normal: Large: Extra Small: Special cases handled in the program: abnormal conditions (unreasonable input):

[Drawing]:

[One brush]:

    1. It is equivalent to understand the position of the two points of "seeking distance". So the range is the same, all <length.
    2. When using Getordefault, you should specify the default type in advance, and take a little notice

[Two brushes]:

[Three brushes]:

[Four brushes]:

[Five brushes]:

[Results of five-minute visual debug]:

[Summary]:

"Distance equal" can be understood as "the same distance appears two times", with HashMap

[Complexity]:time Complexity:o (n^2 double loop ) Space complexity:o (n)

[English data structure or algorithm, why not other data structures or algorithms]:

Do not know how distance means: a[0]-b[0], a[1]-b[1] sum of squares

[Key templating code]:

To find the distance:

 Public int getdistance (intint[] b) {        int dx = a[0]-b[0];         int dy = a[1]-b[1];                 return DX * dx + dy * dy;    }

[Other solutions]:

[Follow up]:

[The problem given by the LC becomes variable]:

[Code Style]:

classSolution { Public intNumberofboomerangs (int[] points) {        //cc        if(Points = =NULL|| Points[0] = =NULL) {            return0; }                //INI HashMapHashmap<integer, integer> map =NewHashmap<>(); intres = 0; //For loop:put into map//Res + = val * (val-1)//. Clear ()         for(inti = 0; i < points.length; i++) {             for(intj = 0; J < Points.length; J + +) {                if(i = = j)Continue; intD =getdistance (Points[i], points[j]); Map.put (d, Map.getordefault (d,0) + 1); }             for(intval:map.values ()) {Res+ = Val * (val-1);        } map.clear (); }        //return res        returnRes; }         Public intGetdistance (int[] A,int[] b) {intDX = a[0]-b[0]; intDY = a[1]-b[1]; returnDX * dx + dy *dy; }}
View Code

447. Number of boomerangs Boomerang array

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