[LeetCode] 836. Rectangle Overlap
An axis-aligned rectangle is represented as a list [x1, y1, x2, y2]
, where (x1, y1)
is the coordinate of its bottom-left corner, and (x2, y2)
is the coordinate of its top-right corner. Its top and bottom edges are parallel to the X-axis, and its left and right edges are parallel to the Y-axis.
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 axis-aligned rectangles rec1
and rec2
, return true
if they overlap, otherwise return false
.
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
Example 3:
Input: rec1 = [0,0,1,1], rec2 = [2,2,3,3] Output: false
Constraints:
rec1.length == 4
rec2.length == 4
-109 <= rec1[i], rec2[i] <= 109
rec1
andrec2
represent a valid rectangle with a non-zero area.
矩形重叠。
矩形以列表 [x1, y1, x2, y2] 的形式表示,其中 (x1, y1) 为左下角的坐标,(x2, y2) 是右上角的坐标。矩形的上下边平行于 x 轴,左右边平行于 y 轴。
如果相交的面积为 正 ,则称两矩形重叠。需要明确的是,只在角或边接触的两个矩形不构成重叠。
给出两个矩形 rec1 和 rec2 。如果它们重叠,返回 true;否则,返回 false 。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/rectangle-overlap
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
这是一道数学题。我给出的思路是去判断两个长方形的可行域是否有交集,如果有,则说明两者有重叠。如何找可行域呢?因为题目说了给的坐标的格式是[x1, y1, x2, y2],前两个数字表示左下角,后两个数字表示右上角。这里计算可行域,我参考了这个帖子。
计算出可行域之后,如果同时满足x1 < x2, y1 < y2则说明有重叠,否则就是没有。
时间O(1)
空间O(1)
Java实现
1 class Solution { 2 public boolean isRectangleOverlap(int[] rec1, int[] rec2) { 3 int x1 = Math.max(rec1[0], rec2[0]); 4 int y1 = Math.max(rec1[1], rec2[1]); 5 int x2 = Math.min(rec1[2], rec2[2]); 6 int y2 = Math.min(rec1[3], rec2[3]); 7 if (x1 < x2 && y1 < y2) { 8 return true; 9 } else { 10 return false; 11 } 12 } 13 }
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