如何将单个对象[]传递给params对象[](How to pass a single object[] to a params object[])

 在标准的计算模型中，这个问题可能没有解决方案明显优于O（n ^ 2）。 
 
 发现三个共线点的问题减少到找到最多点的线的问题，找到三个共线点是3SUM-硬，这意味着在小于O（n ^ 2）时间内解决它将是一个主要的理论结果。 
 
 查看上一个问题 ，找出三个共线点。 
 
 为了您的参考（使用已知的证明），假设我们想回答一个3SUM问题，例如在列表X中找到x，y，z，使得x + y + z = 0。如果我们有一个快速算法的共线点问题，我们可以使用该算法来解决3SUM问题如下。 
 
 对于X中的每个x，创建点（x，x ^ 3）（现在我们假设X的元素是不同的）。 接下来，检查创建点之间是否存在三条共线点。 
 
 要看到这样做，请注意，如果x + y + z = 0，则从x到y的线的斜率为 
 
 （y ^ 3-x ^ 3）/（y-x）= y ^ 2 + yx + x ^ 2 
 
 并且从x到z的线的斜率是 
 
 （z ^ 3-x ^ 3）/（z-x）= z ^ 2 + zx + x ^ 2 =（ - （x + y））^ 2 - （x + y）x + x ^ 2 = 2 + 2xy + y ^ 2 - x ^ 2 - xy + x ^ 2 = y ^ 2 + yx + x ^ 2 
 
 相反，如果从x到y的斜率等于从x到z的斜率 
 
 y ^ 2 + yx + x ^ 2 = z ^ 2 + zx + x ^ 2， 
 
 这意味着 
 
 （y-z）（x + y + z）= 0， 
 
 所以y = z或z = -x-y可以证明减少有效。 
 
 如果X中有重复项，您首先检查x + 2y = 0，对于任何x和重复元素y（使用散列的线性时间或使用排序的O（n lg n）时间），然后在缩小到共线点查找问题。 

There is likely no solution to this problem that is significantly better than O(n^2) in a standard model of computation.
 
The problem of finding three collinear points reduces to the problem of finding the line that goes through the most points, and finding three collinear points is 3SUM-hard, meaning that solving it in less than O(n^2) time would be a major theoretical result.
 
See the previous question on finding three collinear points.
 
For your reference (using the known proof), suppose we want to answer a 3SUM problem such as finding x, y, z in list X such that x + y + z = 0. If we had a fast algorithm for the collinear point problem, we could use that algorithm to solve the 3SUM problem as follows. 
 
For each x in X, create the point (x, x^3) (for now we assume the elements of X are distinct). Next, check whether there exists three collinear points from among the created points.
 
To see that this works, note that if x + y + z = 0 then the slope of the line from x to y is
 
(y^3 - x^3) / (y - x) = y^2 + yx + x^2
 
and the slope of the line from x to z is
 
(z^3 - x^3) / (z - x) = z^2 + zx + x^2 = (-(x + y))^2 - (x + y)x + x^2 = x^2 + 2xy + y^2 - x^2 - xy + x^2 = y^2 + yx + x^2
 
Conversely, if the slope from x to y equals the slope from x to z then
 
y^2 + yx + x^2 = z^2 + zx + x^2,
 
which implies that
 
(y - z) (x + y + z) = 0,
 
so either y = z or z = -x - y as suffices to prove that the reduction is valid.
 
If there are duplicates in X, you first check whether x + 2y = 0 for any x and duplicate element y (in linear time using hashing or O(n lg n) time using sorting), and then remove the duplicates before reducing to the collinear point-finding problem.