Valid Square (M)
题目
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.
题意
给定四个点的坐标,判断这四个点能否组成正方形。
思路
按照x坐标从小到大、y坐标从小到大的优先级进行排序,记此时四个点为a、b、c、d,如果能形成正方形,那么四条边分别是ab、ac、bd、cd。只要判断四条边是否相等,以及对角线是否相等即可(需要先排除所有点在一个位置这一特殊情况)。
代码实现
Java
class Solution {
public boolean validSquare(int[] p1, int[] p2, int[] p3, int[] p4) {
int[][] points = { p1, p2, p3, p4 };
Arrays.sort(points, (p, q) -> p[0] == q[0] ? p[1] - q[1] : p[0] - q[0]);
return calc(points[0], points[1]) != 0
&& calc(points[0], points[1]) == calc(points[0], points[2])
&& calc(points[3], points[1]) == calc(points[3], points[2])
&& calc(points[0], points[1]) == calc(points[3], points[1])
&& calc(points[0], points[3]) == calc(points[1], points[2]);
}
private int calc(int[] p, int[] q) {
int a = p[0] - q[0], b = p[1] - q[1];
return a * a + b * b;
}
}