标签:ted gre nbsp poi fast order LIDS public run
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:
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; } }
标签:ted gre nbsp poi fast order LIDS public run
原文地址:https://www.cnblogs.com/ethanhong/p/10278001.html