1037-有效的回旋镖（Valid Boomerang）

向右看齐 2022-01-23 04:27 196阅读 0赞

**题目描述**  
**中文**

回旋镖定义为一组三个点，这些点各不相同且不在一条直线上。
    
    给出平面上三个点组成的列表，判断这些点是否可以构成回旋镖。

**英文**

A boomerang is a set of 3 points that are all distinct and not in a straight line.
    
    Given a list of three points in the plane, return whether these points are a boomerang.

**示例1**

输入：[[1,1],[2,3],[3,2]]
    	输出：true

**示例2**

输入：[[1,1],[2,2],[3,3]]
    输出：false

**提示**

points.length == 3
    points[i].length == 2
    0 <= points[i][j] <= 100

**解题思路**

*  首先判重。

if((points[0][0] == points[1][0] && points[0][1] == points[1][1])
            		|| (points[0][0] == points[2][0] && points[0][1] == points[2][1])
            		|| (points[1][0] == points[2][0] && points[1][1] == points[2][1])){
            	return false;
            }

*  然后判断是否在一条直线

else {
    			int x = points[2][0],y = points[2][1];
    			int x1 = points[0][0],y1 = points[0][1];
    			int x2 = points[1][0], y2 = points[1][1];
    			if((y-y1)*(x2 - x1) == (y2 - y1)*(x - x1))
    				return false;
    		}

*  然后就没了

**完整代码**

class Solution {
        public boolean isBoomerang(int[][] points) {
            if((points[0][0] == points[1][0] && points[0][1] == points[1][1])
            		|| (points[0][0] == points[2][0] && points[0][1] == points[2][1])
            		|| (points[1][0] == points[2][0] && points[1][1] == points[2][1])){
            	return false;
            }
            else {
    			int x = points[2][0],y = points[2][1];
    			int x1 = points[0][0],y1 = points[0][1];
    			int x2 = points[1][0], y2 = points[1][1];
    			if((y-y1)*(x2 - x1) == (y2 - y1)*(x - x1))
    				return false;
    		}
            return true;
        }
    }