Leetcode: Max Points on a Line - c++编程基础

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
  
    &points) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		unordered_map
   
     mp; int maxNum = 0; for(int i = 0; i < points.size(); i++) { mp.clear(); mp[INT_MIN] = 0; int 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 : (float)(points[j].y - points[i].y)/(points[j].x - points[i].x); mp[k]++; } unordered_map
    
     ::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！

参考：

http://fisherlei.blogspot.com/2013/12/leetcode-max-points-on-line-solution.html