leetcode 149. Max Points on a Line 计算斜率的问题

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

本题就是统计在一条线上的最多的点的数量。

解决方法就是遍历求解，但是主语Java的浮点数的计算精度的问题。

代码如下：

import java.math.BigDecimal;import java.util.HashMap;/*class Point {      int x;      int y;      Point() { x = 0; y = 0; }      Point(int a, int b) { x = a; y = b; }}*//* * 注意Java的浮点数精度 * */public class Solution{    public int maxPoints(Point[] points)    {        if (points.length <= 2)            return points.length;        int max = 2;        for (int i = 0; i < points.length; i++)         {            int pointMax = 1, samePointCount = 0;            HashMap<Double, Integer> slopeCount = new HashMap<Double, Integer>();            for (int j = i + 1; j < points.length; j++)             {                if (points[i].x == points[j].x && points[i].y == points[j].y)                {                     samePointCount++;                     continue;                }                          double k;                if (points[i].x == points[j].x)                    k = Float.POSITIVE_INFINITY;                //这里是为了避免出现-0.0和0.0的问题                else if (points[i].y == points[j].y)                    k = 0.0;                else                {                    //这里是为了避免出现（k,k+1）,(k+1,k+2)和（0,0）的斜率精度问题                    //这里是考虑到了Java的Double的精度问题，所以才是用BigDecimal的问题                    BigDecimal fenziBigDecimal=new BigDecimal(points[i].x-points[j].x);                       BigDecimal fenmuBigDecimal=new BigDecimal(points[i].y-points[j].y);                                 k = fenziBigDecimal.divide(fenmuBigDecimal,20,BigDecimal.ROUND_HALF_DOWN).doubleValue();                }                slopeCount.put(k, slopeCount.getOrDefault(k, 1)+1);                pointMax = Math.max(pointMax, slopeCount.get(k));            }            max = Math.max(max, pointMax+samePointCount);        }        return max;     }}

下面是C++的做法，就是遍历所有的可能性，注意可能出现极端情况，所以这里使用了long double 来计算极端情况的斜率值

代码如下：

#include <iostream>#include <climits>#include <map>#include <vector>#include <algorithm>using namespace std;/*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)     {        if (points.size() <= 2)            return points.size();        int maxRes = 2;        for (int i = 0; i < points.size(); i++)        {            int pointMax = 1, samePoint = 0;            map<long double, int> count;            for (int j = i + 1; j < points.size(); j++)            {                if (points[i].x == points[j].x && points[i].y == points[j].y)                {                    samePoint++;                    continue;                }                long double k = 0;                if (points[i].x == points[j].x)                    k = numeric_limits<double>::max();                else                    k = ((long double)points[i].y - (long double)points[j].y) / ((long double)points[i].x - (long double)points[j].x);                if (count.find(k) == count.end())                    count[k] = 2;                else                    count[k] += 1;                pointMax = max(pointMax,count[k]);            }            maxRes = max(maxRes, pointMax+samePoint);        }        return maxRes;    }};