447.Number of Boomerangs:修订间差异

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创建页面,内容为“=Description= {{LeetCode |id=number-of-boomerangs |no=447 |difficulty=Medium |category=unknown |collection=Top 150 |title=Number of Boomerangs |summary=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).}} Category:Algorithm Category:LeetCode
 
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|title=Number of Boomerangs
|title=Number of Boomerangs
|summary=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).}}
|summary=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).}}
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:
<syntaxhighlight lang="bash">
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]].
</syntaxhighlight>
Example 2:
<syntaxhighlight lang="bash">
Input: points = [[1,1],[2,2],[3,3]]
Output: 2
</syntaxhighlight>
Example 3:
<syntaxhighlight lang="bash">
Input: points = [[1,1]]
Output: 0
</syntaxhighlight>
= Solution =
== Brute force ==
{{Submission|runtime=799ms|memory=40.97MB|rp=5|mp=96.60}}
<syntaxhighlight lang="java">
class Solution {
    public int numberOfBoomerangs(int[][] points) {
        int sum = 0;
        for(int i = 0; i < points.length; i++) {
            for(int j = i + 1; j < points.length; j++) {
                for( int k = j + 1; k < points.length; k++) {
                    if(distance(points, i, j) == distance(points, i, k))
                        sum += 2;
                    if(distance(points, j, i) == distance(points, j, k))
                        sum += 2;
                    if(distance(points, k, i) == distance(points, k, j))
                        sum += 2;
                }
            }
        }
        return sum;
    }
    private static int distance(int[][] points, int i, int j) {
        int l =  points[i][0] - points[j][0];
        int w =  points[i][1] - points[j][1];
        return l * l + w * w;
    }
}
</syntaxhighlight>




[[Category:Algorithm]]
[[Category:Algorithm]]
[[Category:LeetCode]]
[[Category:LeetCode]]

2024年1月17日 (三) 15:53的版本

Description

#447 Number of Boomerangs Medium
unknown Top 150
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).

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

Solution

Brute force

Runtime 799ms
Memory 40.97MB
class Solution {
    public int numberOfBoomerangs(int[][] points) {
        int sum = 0;
        for(int i = 0; i < points.length; i++) {
            for(int j = i + 1; j < points.length; j++) {
                for( int k = j + 1; k < points.length; k++) {
                    if(distance(points, i, j) == distance(points, i, k))
                        sum += 2;
                    if(distance(points, j, i) == distance(points, j, k))
                        sum += 2;
                    if(distance(points, k, i) == distance(points, k, j))
                        sum += 2;
                }
            }
        }
        return sum;
    }

    private static int distance(int[][] points, int i, int j) {
        int l =  points[i][0] - points[j][0];
        int w =  points[i][1] - points[j][1];
        return l * l + w * w;
    }
}