[LeetCode]Rectangle Area

最新推荐文章于 2019-12-17 20:06:59 发布

原创最新推荐文章于 2019-12-17 20:06:59 发布 · 753 阅读

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本文探讨了如何优化计算二维平面上两个矩形总面积的方法，通过简化计算过程提高效率。

Find the total area covered by two rectilinear rectangles in a 2D plane.

Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.

Assume that the total area is never beyond the maximum possible value of int.

对各种case做好处理，分别分析相交图形的长宽，求面积。

class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int l1 = 0;
        int l2 = 0;
        if(A<=E&&G<=C)
            l1 = G-E;
        if(A<=E&&E<=C&&C<=G)
            l1 = C-E;
        if(E<=A&&A<=G&&G<=C)
            l1 = G-A;
        if(E<=A&&C<=G)
            l1 = C-A;
            
        if(B<=F&&H<=D)
            l2 = H-F;
        if(B<=F&&F<=D&&D<=H)
            l2 = D-F;
        if(F<=B&&B<=H&&H<=D)
            l2 = H-B;
        if(F<=B&&D<=H)
            l2 = D-B;
       return (C-A)*(D-B)+(G-E)*(H-F)-l1*l2;
    }
};

这段代码太过冗余，实际上对两个线段求相交时，就是取前两个点的最小值和后两个点的最大值。可以用一行代码解决。

class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
       return (C-A)*(D-B)+(G-E)*(H-F)-max(0,min(C,G)-max(E,A))*max(0,min(D,H)-max(F,B));
    }
};