LeetCode 223 Rectangle Area (rectangular Area)

LeetCode 223 Rectangle Area (rectangular Area)
Translation

Find the total area of the two intersecting rectangles in the two-dimensional plane. Each rectangle defines the coordinates in the lower left corner and the upper right corner. (For example, a rectangle) assumes that the total floor space will never exceed the maximum value of an int.

Original

Analysis

I tried this question the day before yesterday and wrote a bunch of judgments. After all, it was still fruitless ......

Post several other people's solutions ......

int computeArea(int A, int B, int C, int D, int E, int F, int G, int H){    int64_t xmin1 = min( A, C );    int64_t xmax1 = max( A, C );    int64_t ymin1 = min( B, D );    int64_t ymax1 = max( B, D );    int64_t xmin2 = min( E, G );    int64_t xmax2 = max( E, G );    int64_t ymin2 = min( F, H );    int64_t ymax2 = max( F, H );    int64_t xa = min( xmax1, xmax2 ) - max( xmin1, xmin2 );    int64_t ya = min( ymax1, ymax2 ) - max( ymin1, ymin2 );    int64_t z = 0, ca = max( xa, z ) * max( ya, z );    int64_t a1 = (xmax1 - xmin1) * (ymax1 - ymin1);    int64_t a2 = (xmax2 - xmin2) * (ymax2 - ymin2);    return a1 + a2 - ca;}

int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {    int overlap = (min(C,G)-max(A,E))*(min(D,H)-max(B,F));    if ( min(C,G)<=max(A,E) || min(D,H)<=max(B,F) )        overlap = 0;    return (C-A)*(D-B)+(G-E)*(H-F)-overlap;}

int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {    int common = (C <= E || A >= G || B >= H || D <= F) ? 0 : (min(C, G) - max(A, E)) * (min(D, H) - max(B, F));    return (C - A) * (D - B) + (G - E) * (H - F) - common;}