description
There are some points in an XY coordinate system. We use the array coordinates to record their coordinates respectively, where coordinates[i] = [x, y] represents the point with the abscissa x and the ordinate y.
Please judge whether these points are on the same straight line in the coordinate system. If yes, return true, otherwise, return false.
Example 1:
Input: coordinates = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7]]
Output: true
Example 2:
Input: coordinates = [[1,1],[2,2],[3,4],[4,5],[5,6],[7,7]]
Output: false
prompt:
2 <= coordinates.length <= 1000
coordinates[i].length == 2
-10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4
coordinates 中不含重复的点
Source: LeetCode
Link: https://leetcode-cn.com/problems/check-if-it-is-a-straight-line/
Solve
Method two reference https://leetcode-cn.com/problems/check-if-it-is-a-straight-line/solution/san-dian-xiang-chai-zhi-bi-li-chao-99-by -co4x/
struct PairPointsCompare {
bool operator()(const vector<int> &lhs, const vector<int> &rhs) {
if (lhs[0] != rhs[0]) {
return lhs[0] < rhs[0];
}
return lhs[1] < rhs[1];
}
};
class Solution {
public:
// 方法一，暴力解法，对点按照X轴排序后依次对两点求取夹角
bool checkStraightLine_1e(vector<vector<int>> &coordinates) {
std::sort(coordinates.begin(), coordinates.end(), PairPointsCompare());
int l = 0;
int r = coordinates.size() - 1;
double angle = accAngel(coordinates[l], coordinates[r]);
const double DIFF = 1E-6;
while (++l < --r) {
double diff = accAngel(coordinates[l], coordinates[r]) - angle;
if ((diff < -DIFF) || (diff > DIFF)) {
return false;
}
}
if (l == r) {
double diff = accAngel(coordinates[0], coordinates[l]) - angle;
return (diff > -DIFF) && (diff < DIFF);
}
return true;
}
// 方法二，通过斜率相同转换公式，三个点数据进行比较，去除浮点数计算
bool checkStraightLine(vector<vector<int>> &coordinates) {
const int n = coordinates.size();
for (int i = 1; i < n - 1; ++i) {
// 如果只有两个点，一定是一条直线，所以不进入该循环没有任何问题
if (((coordinates[i + 1][1] - coordinates[i][1]) * (coordinates[i][0] - coordinates[i - 1][0])) !=
((coordinates[i][1] - coordinates[i - 1][1]) * (coordinates[i + 1][0] - coordinates[i][0]))) {
return false;
}
}
return true;
}
private:
inline double accAngel(const vector<int> &point1, const vector<int> &point2) {
int xDiff = point2[0] - point1[0];
if (xDiff == 0) {
return 720.0; // 给一个无效夹角值
}
return double(point2[1] - point1[1]) / xDiff;
}
};