• 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;
        }
    }
        
    

      

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  • 原文地址:https://www.cnblogs.com/zlz-ling/p/4035504.html
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