149. Max Points on a Line

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

Example 1:

Input: [[1,1],[2,2],[3,3]]
Output: 3
Explanation:
^
|
|        o
|     o
|  o  
+------------->
0  1  2  3  4

Example 2:

Input: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4
Explanation:
^
|
|  o
|     o        o
|        o
|  o        o
+------------------->
0  1  2  3  4  5  6

 

my code: can't determine the hash value well.

this case can't pass.

[[0,0],[94911151,94911150],[94911152,94911151]]

/**
 * Definition for a point.
 * struct Point {
 *     int x;
 *     int y;
 *     Point() : x(0), y(0) {}
 *     Point(int a, int b) : x(a), y(b) {}
 * };
 */
class Solution {
public:
    int maxPoints(vector<Point>& points) {
        int size = points.size();
        int ans = 0;
        if (size == 0) return 0;
        unordered_map<double, int> mp;
        double k;
        for (int i = 0; i < size; ++i) {
            int num = 0;
            for (int j = i + 1; j < size; ++j) {
                if (points[i].x == points[j].x && points[i].y == points[j].y) {
                    num++;
                    continue;
                }
                if (points[j].x - points[i].x != 0)
                    k = (double)(points[j].y - points[i].y) / (double)(points[j].x - points[i].x);  // how to determine the hash value.
                else k = INT_MAX;
                mp[k]++;
            }
            if (mp[k] == 0) mp[k] = 1, num--;
            for (auto it = mp.begin(); it != mp.end(); ++it) {
                if (it->second > ans) {
                    ans = it->second;
                    ans += num;
                }    
            }
            mp.clear();
        }
        return ans+1;
    }
};

　　

In above test case, when it calculate the slope with [0,0] and [94911151,94911150] it comeback k = 1. So its not safe to store the hash k using the slope. 

Now we have to change our strategy. 

Approach #1: C++.

/**
 * Definition for a point.
 * struct Point {
 *     int x;
 *     int y;
 *     Point() : x(0), y(0) {}
 *     Point(int a, int b) : x(a), y(b) {}
 * };
 */
class Solution {
public:
    int maxPoints(vector<Point>& points) {
        int size = points.size();
        int ans = 0;
        if (size == 0) return 0;
        map<pair<int, int>, int> mp;
        double k;
        int dx, dy;
        int flag;
        for (int i = 0; i < size; ++i) {
            int num = 0;
            for (int j = i + 1; j < size; ++j) {
                if (points[i].x == points[j].x && points[i].y == points[j].y) {
                    num++;
                    continue;
                }
                dx = points[j].x - points[i].x;
                dy = points[j].y - points[i].y;
                flag = gcd(dx, dy);
                mp[{dx/flag, dy/flag}]++;
            }
            ans = max(ans, num);
            for (auto it = mp.begin(); it != mp.end(); ++it) {
                if (it->second + num > ans) {
                    ans = it->second + num;
                }    
            }
            mp.clear();
        }
        return ans+1;
    }
    int gcd (int x, int y) {
        if (y == 0) return x;
        else return gcd(y, x%y);
    }
};

In this case we use the pair of {dx, dy} as the hash key stroed in the map .

 

Approach #2: Java.

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
class Solution {
    public int maxPoints(Point[] points) {
        if (points.length <= 0) return 0;
        if (points.length <= 2) return points.length;
        int result = 0;
        for (int i = 0; i < points.length; ++i) {
            HashMap<Double, Integer> hm = new HashMap<Double, Integer>();
            int samex = 1;
            int samep = 0;
            for (int j = 0; j < points.length; ++j) {
                if (j != i) {
                    if (points[j].x == points[i].x && points[j].y == points[i].y) samep++;
                    if (points[j].x == points[i].x) {
                        samex++;
                        continue;
                    }
                    double k = (double)(points[j].y - points[i].y) / (double)(points[j].x - points[i].x);
                    if (hm.containsKey(k)) hm.put(k, hm.get(k)+1);
                    else hm.put(k, 2);
                    result = Math.max(result, hm.get(k)+samep);
                }
            }
            result = Math.max(result, samex);
        }
        return result;
    }
}

　　

Approach #3: Python.

# Definition for a point.
# class Point(object):
#     def __init__(self, a=0, b=0):
#         self.x = a
#         self.y = b

class Solution(object):
    def maxPoints(self, points):
        """
        :type points: List[Point]
        :rtype: int
        """
        l = len(points)
        m = 0
        for i in range(l):
            dic = {'i' : 1}
            same = 0
            for j in range(i+1, l):
                tx, ty = points[j].x, points[j].y
                if tx == points[i].x and ty == points[i].y:
                    same += 1
                    continue
                if points[i].x == tx: slope = 'i'
                else: slope = (points[i].y - ty) * 1.0 / (points[i].x-tx)
                if slope not in dic: dic[slope] = 1
                dic[slope] += 1
                
            m = max(m, max(dic.values()) + same)
        return m

　　

the python and the java versions have the same question with the first code, but I don't want to correct, because I'm not familiar with these language.XP