这道题简单来说就是雷达扫点的问题。一开始我想入了一种错误思路。就是类似于动雷达的方式然后判断在当前的范围内有多少个点这样。但是问题是对于雷达应该转任意角度都可以，所以遍历这个转的角度并不行
class Solution: def visiblePoints(self, points: List[List[int]], angle: int, location: List[int]) -> int: import math def compute_angle(ori,point): x1,y1 = ori[0],ori[1] x2,y2 = point[0],point[1] dx = x2-x1 dy = y2-y1 angle = math.atan2(dy,dx) angle = angle*180/math.pi if dx<0 and dy<0: angle = 360+angle if dx>0 and dy<0: angle = 360+angle return angle max_count = 0 for d in range(361): count = 0 for point in points: temp_angle = compute_angle(location,point) if d-angle/2<=temp_angle<=d+angle/2 or (point[0]==location[0] and point[1]==location[1]): count += 1 max_count = max(count,max_count) return max_count不过这边有个技巧比较值得注意，就是计算两个点的连线和X轴正半轴的夹角，正确的写法如下：
import math ''' atan2(y, x)是4象限反正切，它的取值不仅取决于正切值y/x，还取决于点 (x, y) 落入哪个象限： 当点(x, y) 落入第一象限时，atan2(y, x)的范围是 0 ~ pi/2; 当点(x, y) 落入第二象限时，atan2(y, x)的范围是 pi/2 ~ pi; 当点(x, y) 落入第三象限时，atan2(y, x)的范围是 －pi～－pi/2; 当点(x, y) 落入第四象限时，atan2(y, x)的范围是 -pi/2～0. ''' def compute_angle(ori,point): x1,y1 = ori[0],ori[1] x2,y2 = point[0],point[1] dx = x2-x1 dy = y2-y1 angle = math.atan2(dy,dx) angle = angle*180/math.pi if (dx<0 and dy<0) or (dx>0 and dy<0): angle = 360+angle return angle ori = [0,0] points = [[2,1],[-2,1],[-2,-1],[2,-1]] for point in points: print(compute_angle(ori,point))正确的思路应该是将点的角度排序，然后滑窗去找能覆盖最多点的区间。因为排序之后只需要找到左右边界，然后计算边界内有多少个点就可以了 这边要注意和原点重合的点需要额外计算，然后因为是环形的sliding window，所以常见的技巧是将原来的数组double一下
class Solution: def visiblePoints(self, points: List[List[int]], angle: int, location: List[int]) -> int: arr, extra = [], 0 xx, yy = location for x, y in points: if x == xx and y == yy: extra += 1 continue arr.append(math.atan2(y - yy, x - xx)) arr.sort() arr = arr + [x + 2.0 * math.pi for x in arr] angle = math.pi * angle / 180 r = ans = 0 for l in range(len(arr)): while r<len(arr) and arr[r]-angle<=arr[l]: r += 1 ans = max(ans,r-l) return ans + extra这边注意滑窗的技巧，遍历左边界，然后右边界滑动寻找下一个不符合条件的点
C++版本 注意一下C++计算角度的函数名相同 pi的表示
class Solution { public: int visiblePoints(vector<vector<int>>& points, int angle, vector<int>& location) { vector<double> arr; int extra = 0; for(auto point : points){ if(point[0] == location[0] && point[1] == location[1]){ extra += 1; continue; } arr.push_back(atan2(point[1]-location[1],point[0]-location[0])); } sort(arr.begin(),arr.end()); int n = arr.size(); for(int i=0;i<n;i++){ arr.push_back(arr[i]+2*M_PI); } int max_point = 0; int l = 0, r = 0; while(r < arr.size()){ while(l < r && arr[r] - arr[l] > angle*M_PI/180.0){ l += 1; } max_point = max(max_point,r-l+1); r += 1; } return max_point + extra; } };