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.

c++版本:

/** 
 * 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) {  
        // IMPORTANT: Please reset any member data you declared, as  
        // the same Solution instance will be reused for each test case.  
        unordered_map<float,int> mp;  
        int maxNum = 0;  
        for(int i = 0; i < points.size(); i++)  
        {  
            mp.clear();  
            mp[INT_MIN] = 0;  
            int duplicate = 1;  
            for(int j = 0; j < points.size(); j++)  
            {  
                if(j == i) continue;  
                if(points[i].x == points[j].x && points[i].y == points[j].y)  
                {  
                    duplicate++;  
                    continue;  
                }  
                float k = points[i].x == points[j].x ? INT_MAX : (float)(points[j].y - points[i].y)/(points[j].x - points[i].x);  
                mp[k]++;  
            }  
            unordered_map<float, int>::iterator it = mp.begin();  
            for(; it != mp.end(); it++)  
                if(it->second + duplicate > maxNum)  
                    maxNum = it->second + duplicate;  
        }  
        return maxNum;  
    }  
};  

  java版本:

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
public class Solution{
    public int maxPoints(Point[] points) {
       if(points.length == 0 || points == null) 
            return 0;
       if(points.length == 1)
           return 1;
        int max = 1;
        for(int i = 0; i < points.length; i++){
            int same = 0;
            int localmax = 1;
            HashMap<Float, Integer> fm = new HashMap<Float, Integer>();
            for(int j = 0; j < points.length; j++) {
                if(i == j)
                   continue;
                   
                if(points[i].x == points[j].x && points[i].y == points[j].y) {
                    same ++;
                    continue;
                }
                
                float slop = ((float)(points[i].x - points[j].x)/(points[i].y - points[j].y));
                
                if(fm.containsKey(slop))
                     fm.put(slop, fm.get(slop) + 1);
                else
                    fm.put(slop, 2);
            } 
                for(Integer value : fm.values())
                   localmax = Math.max(localmax, value);
                     
                   localmax += same;
                   max = Math.max(max,localmax);
        }
        return max;
    }
}
    

  

posted @ 2014-10-19 21:12  zlz~ling  阅读(159)  评论(0编辑  收藏  举报