LeetCode 836. Rectangle Overlap (Java版; Easy)

忘是亡心i 2023-07-16 07:56 21阅读 0赞

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## LeetCode 836. Rectangle Overlap (Java版; Easy) ##

#### 题目描述 ####

A rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) are the coordinates of its 
    bottom-left corner, and (x2, y2) are the coordinates of its top-right corner.
    
    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, return whether they overlap.
    
    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
    Notes:
    
    Both rectangles rec1 and rec2 are lists of 4 integers.
    All coordinates in rectangles will be between -10^9 and 10^9.

#### 第一次做; 核心: 1) 找出不会相交的情况, 然后取反; 满足不想交的情况就不会有重叠了 ####

class Solution { 
        public boolean isRectangleOverlap(int[] rec1, int[] rec2) { 
                return !(rec1[2]<=rec2[0] || rec2[2]<=rec1[0] || rec1[1]>=rec2[3] || rec2[1]>=rec1[3]);
        }
    }

#### 第一次做; 我跑偏了, 计算了重叠矩形的面积, 先找出重叠矩形的四个边界, 再判断四个边界是否有效, 进而计算矩形面积; 这道题不需要计算矩形面积的… ####

class Solution { 
        public boolean isRectangleOverlap(int[] rec1, int[] rec2) { 
                //左边界
                long x1 = Math.max(rec1[0], rec2[0]);
                //右边界
                long x2 = Math.min(rec1[2], rec2[2]);
                //下边界
                long y1 = Math.max(rec1[1], rec2[1]) ;
                //上边界
                long y2 = Math.min(rec1[3], rec2[3]);
    
                //面积
                long res = Math.max(0, x2-x1) * Math.max(0, y2-y1);
    
                return res>0;
        }
    }

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