【leetcode】Max Points on a Line（hard）☆

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

思路：

自己脑子当机了，总是想着斜率和截距都要相同。但实际上三个点是一条直线的话只要它们的斜率相同就可以了，因为用了相同的参照点，截距一定是相同的。

大神的做法：

对每一个点a， 找出所有其他点跟a的连线斜率，相同为同一条线，记录下通过a的点的线上最大的点数。

找出每一个点的最大连线通过的点数。 其中最大的就是在同一条直线上最大的点的数量。

注意可能有相同的点，注意斜率可能为无穷大。

int maxPoints(vector<Point> &points) {
    int result = 0;
    for(int i = 0; i < points.size(); i++){
        int samePoint = 1;
        unordered_map<double, int> map;
        for(int j = i + 1; j < points.size(); j++){
            if(points[i].x == points[j].x && points[i].y == points[j].y){
                samePoint++;
            }
            else if(points[i].x == points[j].x){
                map[INT_MAX]++;
            }
            else{
                double slope = double(points[i].y - points[j].y) / double(points[i].x - points[j].x);
                map[slope]++;
            }
        }
        int localMax = 0;
        for(auto it = map.begin(); it != map.end(); it++){
            localMax = max(localMax, it->second);
        }
        localMax += samePoint;
        result = max(result, localMax);
    }
    return result;
}

First, let's talk about mathematics.

How to determine if three points are on the same line?

The answer is to see if slopes of arbitrary two pairs are the same.

Second, let's see what the minimum time complexity can be.

Definitely, O(n^2). It's because you have to calculate all slopes between any two points.

Then let's go back to the solution of this problem.

In order to make this discussion simpler, let's pick a random point A as an example.

Given point A, we need to calculate all slopes between A and other points. There will be three cases:

1. Some other point is the same as point A.

2. Some other point has the same x coordinate as point A, which will result to a positive infinite slope.

3. General case. We can calculate slope.

We can store all slopes in a hash table. And we find which slope shows up mostly. Then add the number of same points to it. Then we know the maximum number of points on the same line for point A.

We can do the same thing to point B, point C...

Finally, just return the maximum result among point A, point B, point C...