[LeetCode] 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 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：

Max Points on a Line

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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