LC 593. Valid Square

最新推荐文章于 2025-09-11 23:50:21 发布

爱不到要偷

最新推荐文章于 2025-09-11 23:50:21 发布

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文章标签： java runtime

原文链接：http://www.cnblogs.com/ethanhong/p/10278001.html

本文介绍了一种算法，用于判断二维空间中给定的四个点是否能构成一个正方形。通过比较边长和对角线长度，该算法能够快速判断输入点是否满足正方形的条件。

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.

从边入手，如果四个点不重合，那么形成的四条边加两个对角线只可能有两个情况，且对角线要长。

Runtime: 16 ms, faster than 57.63% of Java online submissions for Valid Square.

class Solution {
  public boolean validSquare(int[] p1, int[] p2, int[] p3, int[] p4) {
    int[][] p = {{p1[0],p1[1]},{p2[0],p2[1]},{p3[0],p3[1]},{p4[0],p4[1]}};
    int cnt = 0;
    int[] len = new int[6];
    for(int i=0; i<4; i++){
      for(int j=i+1; j<4 ;j++){
        if(p[i][0] == p[j][0] && p[i][1] == p[j][1]) return false;
        len[cnt++] = (p[i][0]-p[j][0])*(p[i][0]-p[j][0])+(p[i][1]-p[j][1])*(p[i][1]-p[j][1]);
      }
    }
    Arrays.sort(len);
    if(len[0] == len[2] && len[4] == len[5] && len[4] > len[1]){
      return true;
    }
    return false;
  }
}