计算同一条线上的最大点

给定二维平面上的 N 个点作为 (x, y) 坐标对，我们需要找到位于同一条线上的最大点数。

例子：

Input : points[] = {-1, 1}, {0, 0}, {1, 1}, 
                    {2, 2}, {3, 3}, {3, 4} 
Output : 4
Then maximum number of point which lie on same
line are 4, those point are {0, 0}, {1, 1}, {2, 2},
{3, 3}

我们可以通过以下方法解决上面的问题——对于每个点 p，计算它与其他点的斜率，并使用地图记录有多少个点具有相同的斜率，由此我们可以找出与 p 位于同一条线上的点数一点。对于每个点继续做同样的事情并更新迄今为止发现的最大点数。

实施中需要注意的一些事项是：
1) 如果两个点是 (x1, y1) 和 (x2, y2) 那么它们的斜率将是 (y2 – y1) / (x2 – x1) 这可能是一个双精度值并且会导致精度问题。为了摆脱精度问题，我们将斜率视为对 ((y2 – y1), (x2 – x1)) 而不是比率，并在插入地图之前通过它们的 gcd 减少对。在下面的代码点中，垂直或重复的代码点被分开处理。

2) 如果我们使用 c++ 中的 unordered_map 或Java的HashMap 来存储斜率对，那么解的总时间复杂度将为 O(n^2)

C++

/* C/C++ program to find maximum number of point
which lie on same line */
#include 
#include 
  
using namespace std;
  
// method to find maximum colinear point
int maxPointOnSameLine(vector< pair > points)
{
    int N = points.size();
    if (N < 2)
        return N;
  
    int maxPoint = 0;
    int curMax, overlapPoints, verticalPoints;
  
    // here since we are using unordered_map 
    // which is based on hash function 
    //But by default we don't have hash function for pairs
    //so we'll use hash function defined in Boost library
    unordered_map, int,boost::
              hash > > slopeMap;
  
    // looping for each point
    for (int i = 0; i < N; i++)
    {
        curMax = overlapPoints = verticalPoints = 0;
  
        // looping from i + 1 to ignore same pair again
        for (int j = i + 1; j < N; j++)
        {
            // If both point are equal then just
            // increase overlapPoint count
            if (points[i] == points[j])
                overlapPoints++;
  
            // If x co-ordinate is same, then both
            // point are vertical to each other
            else if (points[i].first == points[j].first)
                verticalPoints++;
  
            else
            {
                int yDif = points[j].second - points[i].second;
                int xDif = points[j].first - points[i].first;
                int g = __gcd(xDif, yDif);
  
                // reducing the difference by their gcd
                yDif /= g;
                xDif /= g;
  
                // increasing the frequency of current slope
                // in map
                slopeMap[make_pair(yDif, xDif)]++;
                curMax = max(curMax, slopeMap[make_pair(yDif, xDif)]);
            }
  
            curMax = max(curMax, verticalPoints);
        }
  
        // updating global maximum by current point's maximum
        maxPoint = max(maxPoint, curMax + overlapPoints + 1);
  
        // printf("maximum colinear point 
        // which contains current point 
        // are : %d\n", curMax + overlapPoints + 1);
        slopeMap.clear();
    }
  
    return maxPoint;
}
  
// Driver code
int main()
{
    const int N = 6;
    int arr[N][2] = {{-1, 1}, {0, 0}, {1, 1}, {2, 2},
                    {3, 3}, {3, 4}};
  
    vector< pair > points;
    for (int i = 0; i < N; i++)
        points.push_back(make_pair(arr[i][0], arr[i][1]));
  
    cout << maxPointOnSameLine(points) << endl;
  
    return 0;
}

Python3

def maxPoints(points):
        n=len(points)
  
        # upto two points all points will be part of the line
        if n<3:
            return n
  
        max_val=0
  
        # looping for each point
        for i in points:
  
            # Creating a dictionary for every new
            # point to save memory
            d = {} 
            dups = 0
            cur_max = 0
  
            # pairing with all other points 
            for j in points:
                if i!=j:
                    if j[0]==i[0]: #vertical line
                        slope='inf'
                    else:
                        slope=float(j[1]-i[1])/float(j[0]-i[0])
  
                    # Increasing the frequency of slope and 
                    # updating cur_max for current point(i) 
                    d[slope] = d.get(slope,0)+1
                    cur_max=max(cur_max, d[slope])
  
                # if both points are equal same increase 
                # duplicates count.
                # Please note that this will also increment
                # when we map it with itself.
                # we still do it because we will not have to
                # add the extra one at the end.
                else:
                    dups+=1
  
            max_val=max(max_val, cur_max+dups)
  
        return max_val
  
# Driver code
points = [(-1, 1), (0, 0), (1, 1), (2, 2), (3, 3), (3, 4)]
print(maxPoints(points))

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