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.

思路

关键是浮点数做key不靠谱,struct hash以及 int calcGCD(int a, int b)的写法

代码

/**
 * Definition for a point.
 * struct Point {
 *     int x;
 *     int y;
 *     Point() : x(0), y(0) {}
 *     Point(int a, int b) : x(a), y(b) {}
 * };
 */
struct Key{
    int first;
    int second;
    Key(int f, int s) : first(f), second(s){};
    bool operator==(const Key &other) const{
        return first == other.first
              && second == other.second;
    }
};
namespace std {
template <>
struct hash<Key>{
    size_t operator()(const Key& k) const{
        // Compute individual hash values for first,
        // second and third and combine them using XOR
        // and bit shifting:
        return hash<int>()(k.first)
                ^ (hash<int>()(k.second) << 1);
    }
};
}
class Solution {
private:
    int calcGCD(int a, int b) {
        if (b == 0) //end (divisible, it is the gcd)
            return a;
        return calcGCD(b, a % b);
    }
    Key norm(int a, int b) {
        int gcd = calcGCD(a, b);
        if(gcd == 0) return Key(0,0);//同一个点
        return Key(a/gcd, b/gcd);
    }
public:
    int maxPoints(vector<Point> &points) {
        if(points.empty()) return 0;//不然会Input: []; Output: 1
        unordered_map<Key, int> nSamelines;
        int maxSamelines = 0;
        //每次fix一个点,看其他点和它的共线情况
        for(int i = 0; i < points.size(); i++){
            nSamelines.clear();
            nSamelines.emplace(Key(0,0), 1);
            for(int j = i+1; j < points.size(); j++){
                Key slope = norm(points[i].y-points[j].y, points[i].x-points[j].x);
                //if(slope == Key(0,0)) continue;//得看题意了Input:[(0,0),(0,0)] Output:1 Expected:2
                auto it = nSamelines.find(slope);
                if(it != nSamelines.end()){
                    it->second += 1;
                } else {
                    nSamelines.emplace(slope, 1);
                }
            }
            if(maxSamelines < nSamelines[Key(0,0)]) 
                maxSamelines = nSamelines[Key(0,0)];
            for(auto entry : nSamelines){
                if(!(entry.first == Key(0,0)) && (maxSamelines < entry.second + nSamelines[Key(0,0)])) {
                    maxSamelines = entry.second + nSamelines[Key(0,0)];
                }
            }
        }
        return maxSamelines;
    }
};

posted on 2014-09-20 10:37 小唯THU 阅读(...) 评论(...) 编辑 收藏

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