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.

• 得到兩個長方形的最大覆蓋面積；主要是處理重疊部分

void swapTwo(int *A, int *B) {int tmp = *A;*A = *B;*B = tmp; } void sortFour(int *A, int *C, int *E, int *G) {int tmp = 0;if(*A > *C)swapTwo(A, C);if(*E > *G)swapTwo(E, G);if(*A > *E)swapTwo(A, E);if(*C > *E)swapTwo(C, E);if(*E > *G)swapTwo(E, G); } int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {int total = (C - A) * (D - B) + (G - E) * (H - F);if(A >= G || C <= E || B >= H || D <= F)return total;else{sortFour(&A, &C, &E, &G);sortFour(&B, &D, &F, &H);return total - (E - C) * (F - D);} }

• 如果存在重疊部分，則把橫縱各4個坐標排序，得到中間兩個，然后得到重疊部分的面積

int max(int a, int b) {if(a > b)return a;else return b; } int min(int a, int b) {if(a > b)return b;elsereturn a; } int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {int total = (C - A) * (D - B) + (G - E) * (H - F);if(A >= G || C <= E || B >= H || D <= F)return total;int top = min(D, H);int bottom = max(B, F);int left = max(A, E);int right = min(C, G);return total - (top - bottom) * (right - left); }

• 看起來比上面的代碼簡單，可是速度竟然沒有加快；

轉載于:https://www.cnblogs.com/dylqt/p/4908217.html

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