LintCode 626: Rectangle Overlap (非常容易错的容易题)

本文已参与「新人创作礼」活动，一起开启掘金创作之路。

原文链接：blog.csdn.net/roufoo/arti…

Rectangle Overlap Given two rectangles, find if the given two rectangles overlap or not. Example Given l1 = [0, 8], r1 = [8, 0], l2 = [6, 6], r2 = [10, 0], return true

Given l1 = [0, 8], r1 = [8, 0], l2 = [9, 6], r2 = [10, 0], return ``false`

Notice l1: Top Left coordinate of first rectangle. r1: Bottom Right coordinate of first rectangle. l2: Top Left coordinate of second rectangle. r2: Bottom Right coordinate of second rectangle.

l1 != r2 and l2 != r2

这题容易想糊涂，因为两个矩阵overlap的情况很多。应该反过来思考两个矩阵不overlap的情况。实际上如果一个矩阵在另一个矩阵的上面，或一个矩阵在另一个矩阵的右边，这样两个矩阵一定不重叠，否则就重叠。

/**
 * Definition for a point.
 * struct Point {
 *     int x;
 *     int y;
 *     Point() : x(0), y(0) {}
 *     Point(int a, int b) : x(a), y(b) {}
 * };
 */

class Solution {
public:
    /**
     * @param l1: top-left coordinate of first rectangle
     * @param r1: bottom-right coordinate of first rectangle
     * @param l2: top-left coordinate of second rectangle
     * @param r2: bottom-right coordinate of second rectangle
     * @return: true if they are overlap or false
     */
    bool doOverlap(Point &l1, Point &r1, Point &l2, Point &r2) {
        if ((l2.x > r1.x) || (l1.x > r2.x)) return false;
        if ((r2.y > l1.y) || (r1.y > l2.y)) return false;
        return true;
    }
};