leetcode 836. Rectangle Overlap overlapping rectangles 360 and the spring trick Ali

leetcode 2020 March 1 question daily punch
this question in360 Spring recruit and AliHave encountered

Title:
Rectangular a list [x1, y1, x2, y2 ] expressed in the form, where (x1, y1) coordinates of the lower left corner, (x2, y2) are the coordinates of the upper right corner. If the intersecting area is positive, called two rectangles overlap. To be clear, the two rectangular sides or only at the corners of the contact does not constitute overlap. Given two rectangles, it determines whether they overlap and returns the result.

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

Tip:
two rectangular rec1 rec2 and are given in the form of a list containing four integers.
All coordinates in the rectangle are between 9 and 10 ^ -10 ^ 9.
default x-axis points to the right, y-axis pointing to the default.
You can consider only rectangular case is being put.

Ideas: Python Math question: length and width are respectively intersect

Code:

class Solution(object):
    def isRectangleOverlap(self, rec1, rec2):
        """
        :type rec1: List[int]
        :type rec2: List[int]
        :rtype: bool
        """

        # 长和宽都分别相交
        h1=rec1[2]-rec1[0] #rec1的长
        v1=rec1[3]-rec1[1] #rec1的宽
        h2=rec2[2]-rec2[0] #rec2的长
        v2=rec2[3]-rec2[1] #rec2的宽

        hflag=False
        vflag=False

        if rec1[0]<=rec2[0]:#rec1在rec2左面
            if rec2[0]-rec1[0]<h1:
                hflag=True
        else:#rec1在rec2右面
            if rec1[0]-rec2[0]<h2:
                hflag=True

        if rec1[1]<=rec2[1]:#rec1在rec2下面
            if rec2[1]-rec1[1]<v1:
                vflag=True
        else:#rec1在rec2上面
            if rec1[1]-rec2[1]<v2:
                vflag=True

        return hflag and vflag

This blog is an original work, welcomed the guidance, reproduced, please indicate the source, attach a link to this article, thank you!