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  • 836. Rectangle Overlap 矩形重叠

    [抄题]:

    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

     [暴力解法]:

    时间分析:

    空间分析:

     [优化后]:

    时间分析:

    空间分析:

    [奇葩输出条件]:

    [奇葩corner case]:

    [思维问题]:

    [英文数据结构或算法,为什么不用别的数据结构或算法]:

    [一句话思路]:

    [输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

    [画图]:

    [一刷]:

    [二刷]:

    [三刷]:

    [四刷]:

    [五刷]:

      [五分钟肉眼debug的结果]:

    [总结]:

    [复杂度]:Time complexity: O(n) Space complexity: O(1)

    [算法思想:迭代/递归/分治/贪心]:

    [关键模板化代码]:

    [其他解法]:

    [Follow Up]:

    [LC给出的题目变变变]:

     [代码风格] :

     [是否头一次写此类driver funcion的代码] :

     [潜台词] :

    class Solution {
        public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
            return rec1[0] < rec2[2] && rec1[1] < rec2[3] && rec2[0] < rec1[2] && rec2[1] < rec1[3];
        }
    }
    View Code

     

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  • 原文地址:https://www.cnblogs.com/immiao0319/p/9457699.html
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