牛客 -- leetcode -- max-points-on-a-line

题目描述

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

我的想法：

1、算法设计很简单的题目，但是真正用数据结构实现起来却不容易的一道题。

2、刚开始考虑分三类进行解决，一类横坐标相同，一类纵坐标相同的点，一类斜率点

3、考虑不周，点的坐标可正可负，还想着用数组下标保存，有点蠢

4、思路半路夭折，看了别人的思路。

int maxPoints(vector<Point> &points) {int n = points.size();if (n == 0) return 0;int maxNum = INT_MIN;map<int, vector<Point>> k;set<int> kindK;int shu[10005] = {0}; //shu[i]表示纵坐标相同是i的点的个数int heng[10005] = {0}; //heng[i]表示横坐标相同是i的点的个数int maxX = INT_MIN, maxY = INT_MIN;int minX = INT_MAX, minY = INT_MAX;for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {int ix = points[i].x, iy = points[i].y;maxX = max(ix, maxX);minX = min(ix, minX);maxY = max(maxY, iy);minY = min(iy,minY);int jx = points[j].x, jy = points[j].y;maxX = max(jx, maxX);minX = min(jx, minX);maxY = max(maxY, jy);minY = min(jy,minY);if (ix == jx) {  //在一条竖线上，横坐标相同heng[ix]++;} else if (iy == jy) {shu[iy]++;} else {int xie = (jy - iy) / (jx - ix);kindK.insert(xie);vector<Point> tmp;tmp = k[xie];tmp.push_back(points[i]);tmp.push_back(points[j]);k[xie] = tmp;}}}for(int i = minX; i <= maxX; i++){if(heng[i] == 0) continue;else{maxNum = max(maxNum, heng[i]);}}for(int i = minY; i <= maxY; i++){if(shu[i] == 0) continue;else{maxNum = max(maxNum, shu[i]);}}for(auto x : kindK){vector<Point> tmp = k[x];int b[10005] = {0};int maxB = INT_MIN;for(int i = 0; i < tmp.size(); i++){int bb = tmp[i].y - x*tmp[i].x;b[bb]++;maxB = max(bb,maxB);}for(int i = 0; i <= maxB; i++)}}

牛客网友的思路：

1.分两层循环枚举，第一遍遍历起始点a，第二遍遍历剩下的点

2.分两类，垂直和有斜率的，要注意计算垂直点不为0时，不能重复计算

链接：https://www.nowcoder.com/questionTerminal/bfc691e0100441cdb8ec153f32540be2?f=discussion
来源：牛客网/*** 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) {int size = points.size();if(size == 0)return 0;else if(size == 1)return 1;int ret = 0;for(int i = 0;i<size;i++){int curmax = 1;map<double,int>mp;int vcnt = 0; //垂直点int dup = 0; //重复点for(int j = 0;j<size;j++){if(j!=i){double x1 = points[i].x - points[j].x;double y1 = points[i].y - points[j].y;if(x1 == 0 && y1 == 0){   //重复dup++;}else if(x1 == 0){      //垂直if(vcnt == 0)vcnt = 2;elsevcnt++;curmax = max(vcnt,curmax);}else{double k = y1/x1;          //斜率if(mp[k] == 0)mp[k] = 2;elsemp[k]++;curmax = max(mp[k],curmax);}                   }}ret = max(ret,curmax+dup);           }return ret;}
};

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