[LeetCode]Max Points on a Line
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Question:
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
分析:
任意一条直线都可以表述为
y = ax + b
假设,有两个点(x1,y1), (x2,y2),如果他们都在这条直线上则有
y1 = kx1 +b
y2 = kx2 +b
由此可以得到关系,k = (y2-y1)/(x2-x1)。即如果点c和点a的斜率为k, 而点b和点a的斜率也为k,可以知道点c和点b也在一条线上。
取定一个点points[i], 遍历其他所有节点, 然后统计斜率相同的点数,并求取最大值即可
/** * 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; // for processing duplicate pointint 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 : // key for vertical line (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; }};
注意:
0、points中重复出现的点。
1、int maxNum = 0;
初始化,以防points.size() ==0的情况。
2、mp[INT_MIN] = 0;
保证poins中只有一个结点,还有points中只有重复元素时,mp中没有元素。这两种极端情况。
3、int duplicate = 1;
duplicate记录重复点的数量,初始化为1,是因为要把当前的点points[i]加进去。
4、float k = points[i].x == points[j].x ? INT_MAX : (float)(points[j].y - points[i].y)/(points[j].x - points[i].x);
计算斜率,如果直线和y轴平行,就取INT_MAX,否则就取(float)(points[j].y - points[i].y)/(points[j].x - points[i].x)
一开始把(float)(points[j].y - points[i].y)/(points[j].x - points[i].x)写做(float)((points[j].y - points[i].y)/(points[j].x - points[i].x))一直就不对,后来才想明白,注意注意!
http://blog.csdn.net/doc_sgl/article/details/17103427
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