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[LeetCode]Rectangle Area

时间:2015-06-08 17:26:10      阅读:183      评论:0      收藏:0      [点我收藏+]

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.

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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));
    }
};




[LeetCode]Rectangle Area

原文:http://blog.csdn.net/ciaoliang/article/details/46413825

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