面试题-手撕IOU计算

一般在面试中只会考察二维无旋转矩形框IOU的计算，更复杂的二维框带旋转、三维框等一般不会让写。

二维无旋转矩形框IOU计算

最简单的iou计算， 矩形框，无旋转
2个框的排布可能有下面三种情况，都需要考虑到

代码

#include 
#include 

struct Rectangle {
    float x1, y1, x2, y2; // 左上角和右下角坐标
};

float intersectionArea(const Rectangle& rect1, const Rectangle& rect2) {
    float overlapWidth = std::max(0.0f, std::min(rect1.x2, rect2.x2) - std::max(rect1.x1, rect2.x1));
    float overlapHeight = std::max(0.0f, std::min(rect1.y2, rect2.y2) - std::max(rect1.y1, rect2.y1));

    return overlapWidth * overlapHeight;
}

float calculateIOU(const Rectangle& rect1, const Rectangle& rect2) {
    float area1 = (rect1.x2 - rect1.x1) * (rect1.y2 - rect1.y1);
    float area2 = (rect2.x2 - rect2.x1) * (rect2.y2 - rect2.y1);

    float intersection = intersectionArea(rect1, rect2);
    float unionArea = area1 + area2 - intersection;

    return intersection / unionArea;
}

int main() {
    Rectangle rect1 = {50, 50, 150, 150}; // 示例矩形1
    Rectangle rect2 = {100, 100, 200, 200}; // 示例矩形2

    float iou = calculateIOU(rect1, rect2);
    std::cout << "IOU: " << iou << std::endl;

    return 0;
}