LeetCode--Max Points on a Line

思路：

时间复杂度：O（n^2），挨个遍历。暴力搜索。

但是有几点要注意：第一，map的斜率是float，所以，为了不损失精度，求斜率先把int-》float，再相除；

第二，注意处理为空和为0的情况。为1是返回0还是1，这个和面试官交流，leetcode上面是返回1

/**
 * 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.
        map<float,int> mp;
        int maxNum = 0;
        int i,j;
        for(i = 0 ; i < points.size() ; ++i)
        {
            mp.clear();
            int dup =1;
            mp[INT_MIN] = 0;//为了处理为1的情况
            for(j = 0 ; j < points.size() ; ++j)
            {
                if(i == j)
                    continue;
                if(points[i].x == points[j].x && points[i].y == points[j].y)
                {
                    dup++;
                    continue;
                }
                float k = (points[i].x == points[j].x)?(INT_MAX):(float)(points[i].y-points[j].y)/(points[i].x-points[j].x);
                mp[k]++;
            }
            map<float,int>::iterator it = mp.begin();
            while(it != mp.end())
            {
                if(it->second + dup > maxNum)
                    maxNum = it->second + dup;
                ++it;
            }
        }
        return maxNum;
    }
};