Given n points on a 2D plane, find if there is such a line parallel to y-axis that reflect the given set of points.
Example 1:
Given points = [[1,1],[-1,1]], return true.
Example 2:
Given points = [[1,1],[-1,-1]], return false.
Follow up:
Could you do better than O(n2)?
Hint:
- Find the smallest and largest x-value for all points.
- If there is a line then it should be at y = (minX + maxX) / 2.
- For each point, make sure that it has a reflected point in the opposite side.
Credits:
Special thanks to @memoryless for adding this problem and creating all test cases.
这道题给了我们一堆点，问我们存不存在一条平行于y轴的直线，使得所有的点关于该直线对称。题目中的提示给的相当充分，我们只要按照提示的步骤来做就可以解题了。首先我们找到所有点的横坐标的最大值和最小值，那么二者的平均值就是中间直线的横坐标，然后我们遍历每个点，如果都能找到直线对称的另一个点，则返回true，反之返回false，参见代码如下：
解法一：
class Solution { public: bool isReflected(vector<pair<int, int>>& points) { unordered_map<int, set<int>> m; int mx = INT_MIN, mn = INT_MAX; for (auto a : points) { mx = max(mx, a.first); mn = min(mn, a.first); m[a.first].insert(a.second); } double y = (double)(mx + mn) / 2; for (auto a : points) { int t = 2 * y - a.first; if (!m.count(t) || !m[t].count(a.second)) { return false; } } return true; } };
下面这种解法没有求最大值和最小值，而是把所有的横坐标累加起来，然后求平均数，基本思路都相同，参见代码如下：
解法二：
class Solution { public: bool isReflected(vector<pair<int, int>>& points) { if (points.empty()) return true; set<pair<int, int>> pts; double y = 0; for (auto a : points) { pts.insert(a); y += a.first; } y /= points.size(); for (auto a : pts) { if (!pts.count({y * 2 - a.first, a.second})) { return false; } } return true; } };
本文转自博客园Grandyang的博客，原文链接：直线对称[LeetCode] Line Reflection ，如需转载请自行联系原博主。
