[leetcode] 149. Max Points on a Line

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

解答

找到在同一條直線上的最多的點數。
主要的坑在於，有重複的點，所以遍歷的時候要記錄重複點的個數。
爲了避開使用double，可以保存除數和被除數，但是除以了兩者的最大公約數。用一個map<pair<int,int>, int>記錄同一條直線上的個數。

class Solution {
public:
    int maxPoints(vector<vector<int>>& points) {
        if(points.empty() || points[0].empty())
            return 0;
        int res = 0;
        int num = points.size();
        for(int i=0;i<num;i++){
            map<pair<int,int>,int> m;
            int dup = 1;
            for(int j = i+1;j<num;j++){
                if(points[i][0] == points[j][0] && points[i][1] == points[j][1]) {
                    dup ++;
                    continue;
                }
                int dx = points[i][0] - points[j][0];
                int dy = points[i][1] - points[j][1];
                int d = gcd(dx, dy);
                ++m[{dx / d, dy / d}];
            }
            res = max(res, dup);
            for(auto it = m.begin(); it != m.end(); ++it)
                res = max(res, it->second + dup);
        }
        return res;
    }
    
    int gcd(int a,int b){
        return b == 0 ? a : gcd(b,a%b);
    }
};

參考

https://www.cnblogs.com/grandyang/p/4579693.html