[LintCode] 626 Rectangle Overlap 解题报告

Description
Given two rectangles, find if the given two rectangles overlap or not.

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


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


思路
如果长方形1在长方形2的左边或者右边，我们不需要考虑y坐标，直接返回false。
   ______        _________
   |            |      |                  |
   |______|      |_________|
------------------------------------>

如果长方形1在长方形2的上面活着下面，我们也不需要考虑x坐标，直接返回false。
^
|    _____
|    |        |
|    |____|
|
|      _____
|      |        |
|      |____|
|

其余所有情况，一定会有重叠。


Code

/**
 * Definition for a point.
 * class Point {
 *     public int x, y;
 *     public Point() { x = 0; y = 0; }
 *     public Point(int a, int b) { x = a; y = b; }
 * }
 */

public class Solution {
    /**
     * @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
     */
    public boolean doOverlap(Point l1, Point r1, Point l2, Point r2) {
        // Write your code here
        
        if (l1.x > r2.x || l2.x > r1.x) {
            return false;
        }
        
        if (r1.y > l2.y || r2.y > l1.y) {
            return false;
        }
        
        return true;
    }
}