Leetcode每日一题 —— 836. 矩形重叠

class Solution:

    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:

        return (

            rec1[0] < rec2[2] and

            rec2[0] < rec1[2] and

            rec1[1] < rec2[3] and

            rec2[1] < rec1[3]

        )

非不重叠

class Solution {

public:

    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {

        return !(rec1[0] >= rec2[2] || rec1[2] <= rec2[0] || rec1[1] >= rec2[3] || rec1[3] <= rec2[1]);

    }

};

反向思考，什么时候不行。

class Solution {

public:

    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {

        // 相交面积不为 0，即为重叠

        // 正向思考写起来好麻烦，反向思考什么时候不交叠

        int r1x1=rec1[0],r1y1=rec1[1],r1x2=rec1[2],r1y2=rec1[3];

        int r2x1=rec2[0],r2y1=rec2[1],r2x2=rec2[2],r2y2=rec2[3];

        return !(r1x2<=r2x1||r1x1>=r2x2||r1y2<=r2y1||r1y1>=r2y2);

    }

};

转化为两个正交维度上各自的区间交集不能为空

class Solution:

    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:

        x11, y11, x12, y12 = rec1

        x21, y21, x22, y22 = rec2



        return min(x12, x22) - max(x11, x21) > 0 and min(y12, y22) - max(y11, y21) > 0

• 从一维的区间重叠入手,判断x轴区间和y轴的区间同时重叠就行了

class Solution {

public:

    bool check(int l1,int r1,int l2,int r2)

    {

        if(l1 <= l2) return l2 < r1;

        else return check(l2,r2,l1,r1);

    }



    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) 

    {

        return check(rec1[0],rec1[2],rec2[0],rec2[2]) && check(rec1[1],rec1[3],rec2[1],rec2[3]);

        

    }

};

相交为正所以rec2的左面要在rec1的右面的左侧，rec2的右面要在rec1左面的右侧，上下面同理

public class Solution {

    public bool IsRectangleOverlap(int[] rec1, int[] rec2) {

        return rec2[0] < rec1[2] && rec2[2] > rec1[0] && rec2[1] < rec1[3] && rec2[3] > rec1[1];

    }

}

简单题简单做~

class Solution {

public:

    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {

        int xmn = max(rec1[0], rec2[0]), xmx = min(rec1[2], rec2[2]);

        int ymn = max(rec1[1], rec2[1]), ymx = min(rec1[3], rec2[3]);

        return xmn < xmx && ymn < ymx;

    }

};