836. Rectangle Overlap

中文文档

Description

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 and rec2 represent a valid rectangle with a non-zero area.

Solutions

Solution 1: Determine Non-Overlap Cases

Let the coordinates of rectangle $\text{rec1}$ be $(x_1, y_1, x_2, y_2)$, and the coordinates of rectangle $\text{rec2}$ be $(x_3, y_3, x_4, y_4)$.

The rectangles $\text{rec1}$ and $\text{rec2}$ do not overlap if any of the following conditions are met:

  • $y_3 \geq y_2$: $\text{rec2}$ is above $\text{rec1}$;

  • $y_4 \leq y_1$: $\text{rec2}$ is below $\text{rec1}$;

  • $x_3 \geq x_2$: $\text{rec2}$ is to the right of $\text{rec1}$;

  • $x_4 \leq x_1$: $\text{rec2}$ is to the left of $\text{rec1}$.

If none of the above conditions are met, the rectangles $\text{rec1}$ and $\text{rec2}$ overlap.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

Python3

class Solution:
    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
        x1, y1, x2, y2 = rec1
        x3, y3, x4, y4 = rec2
        return not (y3 >= y2 or y4 <= y1 or x3 >= x2 or x4 <= x1)

Java

class Solution {
    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
    }
}

C++

class Solution {
public:
    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
    }
};

Go

func isRectangleOverlap(rec1 []int, rec2 []int) bool {
	x1, y1, x2, y2 := rec1[0], rec1[1], rec1[2], rec1[3]
	x3, y3, x4, y4 := rec2[0], rec2[1], rec2[2], rec2[3]
	return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1)
}

TypeScript

function isRectangleOverlap(rec1: number[], rec2: number[]): boolean {
    const [x1, y1, x2, y2] = rec1;
    const [x3, y3, x4, y4] = rec2;
    return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
}