Max Points on a Line(直线上最多的点数)

给定一个二维平面,平面上有 个点,求最多有多少个点在同一条直线上。

示例 1:

输入: [[1,1],[2,2],[3,3]]
输出: 3
解释:
^
|
|        o
|     o
|  o  
+------------->
0  1  2  3  4

示例 2:

输入: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
输出: 4
解释:
^
|
|  o
|     o        o
|        o
|  o        o
+------------------->
0  1  2  3  4  5  6

看到此题,第一想法是用一个数据结构来表示某一条直线,直线的表达式有y = ax + b ,那提取出 a和b是不是就可以了,写完发现还有x = 1这种直线。然后想通过hashmap来保存直线对应的点数,如果参数a和b是个1/3这种值,由于精度问题,算出来的两个直线的数据结构的hash值就不一样,hashmap就是认为是两个key。 最后无奈换成分数表达式 y = (a1/a2)*x + b1/b2; b1/b2 = (a2*y - a1*x)/a2;这样4个int变量,就可以精确表示一条直线.当然,分数需要进行约分!
struct FPoint {
    int a1;
    int a2;
    int b1;
    int b2;
    FPoint() : a1(0), a2(0), b1(0), b2(0) {}
    FPoint(int _a1, int _a2)
    {
        if (_a1 * _a2 > 0)
        {
            a1 = abs(_a1);
            a2 = abs(_a2);
        }
        else
        {
            a1 = -1 * abs(_a1);
            a2 = abs(_a2);
        }
        b1 = 0;
        b2 = 1;
    }
    bool Contain(Point p)const
    {
        long long int t1 = 1L, t2 = 1L;
        t1 = t1 * p.y * (a2*b2);
        t2 = t2 * a1*b2*p.x + b1*a2;
        return a2 == 0 ? p.x == b1 / b2 : t1 == t2;
    }
    void Normal()
    {
        if (a1 == 0 && a2 == 0)
        {
        }
        else if (a1 == 0)
            a2 = 1;
        else if (a2 == 0)
            a1 = 1;
        else
        {
            int s = a1*a2 > 0 ? 1 : -1;
            a1 = abs(a1);
            a2 = abs(a2);
            int _gcd = GCD(a1, a2);
            a1 = s * a1 / _gcd;
            a2 = a2 / _gcd;

            if (b1 == 0)
                b2 = 1;
            else
            {
                s = b1*b2 > 0 ? 1 : -1;
                b1 = abs(b1);
                b2 = abs(b2);
                _gcd = GCD(b1, b2);
                b1 = s * b1 / _gcd;
                b2 = b2 / _gcd;
            }
        }

    }
  //最大公约数
int GCD(int a, int b){ int m = a, n = b, r; if (m < n){ int temp = m; m = n; n = temp; } r = m % n; while (r){ m = n; n = r; r = m % n; } return n; } }; struct RecordHash { size_t operator()(const FPoint& f) const{ return hash<int>()(f.a1) ^ hash<int>()(f.a2) ^ hash<int>()(f.b1) ^ hash<int>()(f.b2); } }; struct RecordCmp { bool operator()(const FPoint& f1, const FPoint& f2) const{ return f1.a1 == f2.a1 && f1.a2 == f2.a2 &&f1.b1 == f2.b1&&f1.b2 == f2.b2; } }; class Solution { public: FPoint GetPoint(Point a, Point b) { FPoint f(b.y - a.y, b.x - a.x ); if (f.a2 == 0) { f.b1 = a.x; f.b2 = 1; } else { f.b1 = (f.a2*a.y - f.a1*a.x); f.b2 = f.a2; } return f; } int maxPoints(vector<Point>& points) { if (points.size() <= 1) { return points.size(); } unordered_set<int> index_set; unordered_map<FPoint, int, RecordHash, RecordCmp> line_map; int max_point = 0; for (int i = 0; i < points.size(); i++) { unordered_map<FPoint, int, RecordHash, RecordCmp>::iterator itr = line_map.begin(); for (; itr != line_map.end(); itr++) { if (itr->first.Contain(points[i])) { itr->second++; max_point = max(max_point, itr->second); if (index_set.count(i) == 0)index_set.insert(i); } } for (int r = 0; r < points.size() ; r++) { if (r != i && index_set.count(r) == 0) { FPoint f = GetPoint(points[i], points[r]); f.Normal(); if (line_map[f] == 0) { line_map[f]++; } if (index_set.count(i) == 0) { index_set.insert(i); } max_point = max(max_point, line_map[f]); } } } return max_point; } };

 

posted @ 2018-09-27 16:07  newShit  阅读(427)  评论(0编辑  收藏