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     写一个方法做上述判断，可以很快求出两个矩形交集的height和width 

// ACclass Solution {public:    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {        int AreaA = (C-A) * (D-B);        int AreaB = (G-E) * (H-F);        int intersectionWidth = getIntersection(A, C, E, G);        int intersectionHeight = getIntersection(B, D, F, H);        return AreaA + AreaB - intersectionWidth * intersectionHeight;    }    int getIntersection(int A, int B, int C, int D){        int ret = 0;        if (C >= A && C <= B){            ret = min(B-C, D-C);        }else if (D >= A && D <= B){            ret = D - A;        }else if (C < A && D > B){            ret = B - A;        }        return ret;    }};

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