[LeetCode] 593. Valid Square
grandyang opened this issue · comments
Given the coordinates of four points in 2D space, return whether the four points could construct a square.
The coordinate (x,y) of a point is represented by an integer array with two integers.
Example:
Input: p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,1]
Output: True
Note:
- All the input integers are in the range [-10000, 10000].
- A valid square has four equal sides with positive length and four equal angles (90-degree angles).
- Input points have no order.
这道题给了我们四个点,让验证这四个点是否能组成一个正方形,刚开始博主考虑的方法是想判断四个角是否是直角,但即便四个角都是直角,也不能说明一定就是正方形,还有可能是矩形。还得判断各边是否相等。其实这里可以仅通过边的关系的来判断是否是正方形,根据初中几何的知识可以知道正方形的四条边相等,两条对角线相等,满足这两个条件的四边形一定是正方形。那么这样就好办了,只需要对四个点,两两之间算距离,如果计算出某两个点之间距离为0,说明两点重合了,直接返回 false,如果不为0,那么就建立距离和其出现次数之间的映射,最后如果我们只得到了两个不同的距离长度,那么就说明是正方形了,参见代码如下:
解法一:
class Solution {
public:
bool validSquare(vector<int>& p1, vector<int>& p2, vector<int>& p3, vector<int>& p4) {
unordered_map<int, int> m;
vector<vector<int>> v{p1, p2, p3, p4};
for (int i = 0; i < 4; ++i) {
for (int j = i + 1; j < 4; ++j) {
int x1 = v[i][0], y1 = v[i][1], x2 = v[j][0], y2 = v[j][1];
int dist = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
if (dist == 0) return false;
++m[dist];
}
}
return m.size() == 2;
}
};
我们其实不用建立映射,直接用个集合 HashSet 来放距离就行了,如果最后集合中不存在0,且里面只有两个数的时候,说明是正方形,参见代码如下:
解法二:
class Solution {
public:
bool validSquare(vector<int>& p1, vector<int>& p2, vector<int>& p3, vector<int>& p4) {
unordered_set<int> s{d(p1, p2), d(p1, p3), d(p1, p4), d(p2, p3), d(p2, p4), d(p3, p4)};
return !s.count(0) && s.size() == 2;
}
int d(vector<int>& p1, vector<int>& p2) {
return (p1[0] - p2[0]) * (p1[0] - p2[0]) + (p1[1] - p2[1]) * (p1[1] - p2[1]);
}
};
参考资料:
https://leetcode.com/problems/valid-square/
https://leetcode.com/problems/valid-square/discuss/103442/c-3-lines-unordered_set
请问m[dist]++代表着什么?