力扣每日一题2022-06-08简单题：有效的回旋镖

---

题目描述

有效的回旋镖

---

思路

向量叉乘

从points[0]开始，分别计算指向points[1]和points[2]的向量v1、v2。三点各不相同且不在一条直线上等价于v1v2两个向量的叉乘结果不为0。

Python实现

class Solution:
    def isBoomerang(self, points: List[List[int]]) -> bool:
        v1 = (points[1][0] - points[0][0], points[1][1] - points[0][1])
        v2 = (points[2][0] - points[0][0], points[2][1] - points[0][1])
        return v1[0] * v2[1] - v1[1] * v2[0] != 0

Java实现

class Solution {
    public boolean isBoomerang(int[][] points) {
        int[] v1 = {points[1][0] - points[0][0], points[1][1] - points[0][1]};
        int[] v2 = {points[2][0] - points[0][0], points[2][1] - points[0][1]};
        return v1[0] * v2[1] - v1[1] * v2[0] != 0;
    }
}

C++实现

class Solution {
public:
    bool isBoomerang(vector<vector<int>>& points) {
        vector<int> v1 = {points[1][0] - points[0][0], points[1][1] - points[0][1]};
        vector<int> v2 = {points[2][0] - points[0][0], points[2][1] - points[0][1]};
        return v1[0] * v2[1] - v1[1] * v2[0] != 0;
    }
};