[LeetCode] 356. Line Reflection 直线对称

Given n points on a 2D plane, find if there is such a line parallel to y-axis that reflect the given points.

Example 1:

Input: [[1,1],[-1,1]]
Output: true

Example 2:

Input: [[1,1],[-1,-1]]
Output: false

Follow up:
Could you do better than O(n2) ?

Hint:

1. Find the smallest and largest x-value for all points.
2. If there is a line then it should be at y = (minX + maxX) / 2.
3. 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：

Max Points on a Line

参考资料：

https://leetcode.com/problems/line-reflection/description/

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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