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; } };
类似题目:
参考资料:
https://leetcode.com/discuss/107661/48-ms-short-c-solution
https://leetcode.com/discuss/107761/group-by-y-then-sort-by-x-17ms
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