836. Rectangle Overlap ——weekly contest 85

Rectangle Overlap

A rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) are the coordinates of its bottom-left corner, and (x2, y2) are the coordinates of its top-right corner.

Two rectangles overlap if the area of their intersection is positive.  To be clear, two rectangles that only touch at the corner or edges do not overlap.

Given two rectangles, return whether they overlap.

Example 1:

Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
Output: true

Example 2:

Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
Output: false

Notes:

1. Both rectangles rec1 and rec2 are lists of 4 integers.
2. All coordinates in rectangles will be between -10^9 and 10^9.

 1 class Solution {
 2 public:
 3     bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
 4         vector<int> xx,yy;
 5         xx.push_back(rec1[0]);
 6         xx.push_back(rec1[2]);
 7         xx.push_back(rec2[0]);
 8         xx.push_back(rec2[2]);
 9         yy.push_back(rec1[1]);
10         yy.push_back(rec1[3]);
11         yy.push_back(rec2[1]);
12         yy.push_back(rec2[3]);
13         std::sort(xx.begin(),xx.end());
14         std::sort(yy.begin(),yy.end());
15         for(int x = 0; x < 4;x++){
16             for(int y = 0;y < 4;y++){
17                 if(2*xx[x]+1 < 2*rec2[2]&&2*xx[x]+1 > 2*rec2[0]&&2*yy[y]+1 > 2*rec2[1]&&2*yy[y]+1 < 2*rec2[3]){
18                     if(2*xx[x]+1 < 2*rec1[2]&&2*xx[x]+1 > 2*rec1[0]&&2*yy[y]+1 > 2*rec1[1]&&2*yy[y]+1 < 2*rec1[3]){
19                         return true;
20                     }
21                  }
22             }
23         }
24         return false;
25     }
26 };

A easier solution:

1 bool isRectangleOverlap(vector<int>& a, vector<int>& b) {
2     return !(a[2] <= b[0] || b[2] <= a[0] || a[3] <= b[1] || b[3] <= a[1]);
3 }

原文地址：https://www.cnblogs.com/jinjin-2018/p/9065112.html