Max Points on a Line

最新推荐文章于 2018-07-30 23:28:01 发布

转载最新推荐文章于 2018-07-30 23:28:01 发布 · 66 阅读

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原文链接：http://www.cnblogs.com/winscoder/p/3473032.html

本文介绍了一种算法，用于找出二维平面上最多有多少个点位于同一直线上。通过遍历每两个不同的点并检查其它点是否在这条直线上，该算法的时间复杂度为O(n^3)。

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

Analysis:

　　This approach cost time complexity O(n^3).

　　1. If points less than 2 or all points locate at the same position, just output the points number.

　　2. Find two points with different coordinates and check how many points other than these two are in this line.

　　3. Traverse all the points. Keep updating the max points number.

Code:

/**
 * 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:
    bool allSamePoints(vector<Point> &points){
        for(int i=1;i<points.size();i++)
            if(points[i].x==points[i-1].x && points[i].y==points[i-1].y) return true;
            else  return false;
    }
    bool sameLine(Point i, Point j, Point k){
        return ((i.y-j.y)*(i.x-k.x)-(i.y-k.y)*(i.x-j.x))==0;
    }
    int maxPoints(vector<Point> &points) {
        if(points.size()<=1||allSamePoints(points))
            return points.size();
        int maxPoints = 2;    
        for(int i=0;i<points.size();i++){
            for(int j=i+1;j<points.size();j++){
                if(points[i].x!=points[j].x||points[i].y!=points[j].y){
                    int n=2;
                    for(int k=0;k<points.size();k++){
                        if(k!=i&&k!=j&&sameLine(points[i],points[j],points[k]))
                            n++;
                    }
                    maxPoints=max(maxPoints,n);
                }
            }  
        }
        return maxPoints;
    }
};

转载于:https://www.cnblogs.com/winscoder/p/3473032.html