快速矩形到矩形交集-C/C++问题

Fast rectangle to rectangle intersection(快速矩形到矩形交集)

本文介绍了快速矩形到矩形交集的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

测试两个矩形是否相交的快速方法是什么?

What's a fast way to test if 2 rectangles are intersecting?

网上一搜就找到了这个单行代码(WOOT！)，但我不明白如何用Javascript编写它，它似乎是用一种古老的C++形式编写的.

A search on the internet came up with this one-liner (WOOT!), but I don't understand how to write it in Javascript, it seems to be written in an ancient form of C++.

struct
{
    LONG    left;
    LONG    top;
    LONG    right;
    LONG    bottom;
} RECT; 

bool IntersectRect(const RECT * r1, const RECT * r2)
{
    return ! ( r2->left > r1->right
        || r2->right < r1->left
        || r2->top > r1->bottom
        || r2->bottom < r1->top
        );
}

推荐答案

这就是将代码转换为 JavaScript 的方式.请注意，正如评论所建议的那样，您的代码和文章中的代码有错别字.具体 r2->right left 应该是 r2->right <;r1->left 和 r2->bottom top 应该是 r2->bottom r1->top 使函数起作用.


This is how that code can be translated to JavaScript. Note that there is a typo in your code, and in that of the article, as the comments have suggested. Specifically r2->right left should be r2->right < r1->left and r2->bottom top should be r2->bottom < r1->top for the function to work.
function intersectRect(r1, r2) {
  return !(r2.left > r1.right || 
           r2.right < r1.left || 
           r2.top > r1.bottom ||
           r2.bottom < r1.top);
}

测试用例:
var rectA = {
  left:   10,
  top:    10,
  right:  30,
  bottom: 30
};

var rectB = {
  left:   20,
  top:    20,
  right:  50,
  bottom: 50
};

var rectC = {
  left:   70,
  top:    70,
  right:  90,
  bottom: 90
};

intersectRect(rectA, rectB);  // returns true
intersectRect(rectA, rectC);  // returns false


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