Force deduction solution summary 1037- effective boomerang

Directory links ：

Force buckle programming problem - The solution sums up _ Share + Record -CSDN Blog

GitHub Synchronous question brushing items ：

https://github.com/September26/java-algorithms

Original link ： Power button

describe ：

Given an array  points , among  points[i] = [xi, yi]  Express X-Y A point on the plane , If these points form a   Boomerang   Then return to  true .

Boomerang   Defined as a set of three points , These points   Each are not identical   And   Not in a straight line  .

Example 1：

Input ：points = [[1,1],[2,3],[3,2]]
Output ：true
Example 2：

Input ：points = [[1,1],[2,2],[3,3]]
Output ：false
 

Tips ：

points.length == 3
points[i].length == 2
0 <= xi, yi <= 100

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/valid-boomerang
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Their thinking ：

*  Their thinking ：
*  First, compare the three points to see if they overlap .
*  Then judge whether all three points are X Same axial direction .
*  Finally, the three points are subtracted by two ,Y Axis difference divided by X Axis difference , See if it's the same .

Code ：

    public boolean isBoomerang(int[][] points) {
        int[] point0 = points[0];
        int[] point1 = points[1];
        int[] point2 = points[2];
        if (point0[0] == point1[0] && point0[1] == point1[1]) {
            return false;
        }
        if (point0[0] == point2[0] && point0[1] == point2[1]) {
            return false;
        }
        if (point1[0] == point2[0] && point1[1] == point2[1]) {
            return false;
        }
        if (point0[0] == point1[0] || point0[0] == point2[0]) {
            return point0[0] != point1[0] || point0[0] != point2[0];
        }
        return (double)(point0[1] - point1[1]) / (double)(point0[0] - point1[0]) != (double)(point0[1] - point2[1]) / (double)(point0[0] - point2[0]);
    }