实用指南：LeetCode //C - 836. Rectangle Overlap

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, returntrue 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
• $-10^9<=rec1[i],rec2[i]<=10^9$
• rec1 and rec2 represent a valid rectangle with a non-zero area.

From: LeetCode
Link: 836. Rectangle Overlap

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Solution:

Ideas:

• The rectangles do not overlap if:

  • One is completely to the left/right of the other.

  • One is completely above/below the other.

• We use that logic to check for a “no overlap” condition.

• If none of those are true, the rectangles must overlap, so we return true.

Code:

bool isRectangleOverlap(int* rec1, int rec1Size, int* rec2, int rec2Size) {
// rec1 = [x1, y1, x2, y2]
// rec2 = [x1, y1, x2, y2]
// Check if there is **no overlap**
if (rec1[2] <= rec2[0] || // rec1 is to the left of rec2
rec1[0] >= rec2[2] || // rec1 is to the right of rec2
rec1[3] <= rec2[1] || // rec1 is below rec2
rec1[1] >= rec2[3]) // rec1 is above rec2
{
return false;
}
// If none of the above, rectangles overlap
return true;
}