给定两个矩形,分别以四个数字 [x1,y1,x2,y2] 表示,其中 (x1,y1) 表示矩形左下角,(x2,y2) 表示矩形右上角,矩形的上下边平行于 x 轴,左右边平行于 y 轴。
如果两个矩形相交的面积为正,则两矩形重叠。如果重叠则输出 true ,否则输出 false
数据范围:矩形的四个角坐标满足
[0,0,2,2],[2,2,4,4]
false
[0,0,2,2],[1,1,3,3]
true
package main //import "fmt" /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param r1 int整型一维数组 * @param r2 int整型一维数组 * @return bool布尔型 */ func overlapRec( r1 []int , r2 []int ) bool { if r1[2]<=r2[0]||r1[0]>=r2[2]||r1[3]<=r2[1]||r1[1]>=r2[3]{ return false } return true }
class Solution { public: bool overlapRec(vector<int>& r1, vector<int>& r2) { double half_h_1 = (r1[3] - r1[1]) / 2; double half_x_1 = (r1[2] - r1[0]) / 2; double half_h_2 = (r2[3] - r2[1]) / 2; double half_x_2 = (r2[2] - r2[0]) / 2; double dis_x = abs(r2[2] + r2[0] - r1[2] - r1[0]) / 2; double dis_h = abs(r2[3] + r2[1] - r1[3] - r1[1]) / 2; if (dis_h < half_h_1 + half_h_2 && dis_x < half_x_1 + half_x_2) { return true; } else { return false; } } };
# -*- coding: utf-8 -*- # # 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 # # # @param r1 int整型一维数组 # @param r2 int整型一维数组 # @return bool布尔型 # class Solution: """ 题目: https://www.nowcoder.com/practice/53acad54f18444a9b714acd75115d817?tpId=196&tqId=40505&rp=1&ru=/exam/oj&qru=/exam/oj&sourceUrl=%2Fexam%2Foj%3Fpage%3D8%26tab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D196&difficulty=undefined&judgeStatus=undefined&tags=&title= 算法: 判断矩形是否重叠,我们根据矩形位置,分为左右两个矩形;对于左右矩形是否重叠,实际上就是判断左矩形的右下角和右矩形的左上角是否重叠,或者左矩形的右上角与右矩形的左下角是否重叠 复杂度: 时间复杂度:O(1) 空间复杂度:O(1) """ def overlapRec(self, r1, r2): # write code here if r1[0] > r2[0]: r1, r2 = r2, r1 # print r1, r2 RU1 = (r1[2], r1[3]) # r1的右上角 RD1 = (r1[2], r1[1]) # r1的右下角 LU2 = (r2[0], r2[3]) # r2的左上角 LD2 = (r2[0], r2[1]) # r2的左下角 return RU1[0] > LD2[0] and RU1[1] < LD2[1]&nbs***bsp;\ RD1[0] > LU2[0] and RD1[1] < LU2[1] if __name__ == "__main__": sol = Solution() # r1, r2 = [0, 0, 2, 2], [2, 2, 4, 4] # r1, r2 = [0, 0, 2, 2], [1, 1, 3, 3] r1, r2 = [2, -14, 14, -10], [0, 5, 3, 11] res = sol.overlapRec(r1, r2) print res