[leetcode] 223. Rectangle Area

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.

这道题给出两个矩形，计算他们覆盖的总面积，题目难度为easy。

很容易想到，计算两个矩形的总面积然后减去重叠的部分就得到了总面积，关键在于重叠部分面积的计算。具体代码：

class Solution {
public:
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
int area = 0;
area += (C - A) * (D - B);
area += (G - E) * (H - F);
if(C<=E || A>=G || D<=F || B>=H) return area;
else return area - (min(C, G) - max(A, E)) * (min(D, H) - max(B, F));
}
};

查看别人代码时，看到一条讨论标题是“If
you want to laugh, look at my solution”，不得不承认人才确实多，忍不住笑了，给他点一个大大的赞，代码：

int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
int area1 = (D - B)*(C - A);
int area2 = (H - F)*(G - E);
int area3;
if (area1 == 0) {
return area2;
}
if (area2 == 0) {
return area1;
}
if ((A == D) && (B == F) && (C == G) && (D == H)) {
return area1;
}
if ((E >= C) | (G <= A) | (H <= B) | (D <= F)) {    //not overlapping
return (area1 + area2);
}
if (((G - E) <= (C - A)) && ((H - F) <= (D - B)) && (E >= A) && (F >= B) && (G <= C) && (D >= H)) {                        //rect2 is inside rect1
return area1;
}
if (((C - A) <= (G - E)) && ((D - B) <= (H - F)) && (E <= A) && (B >= F) && (G >= C) && (H >= D)) {                        //rect1 is inside rect2
return area2;
}
if ((F >= B) && (E >= A) && (G >= C) && (H >= D)) {                       //overlapping upper right corner
area3 = (C - E)*(D - F);
}
else if ((F >= B) && (E <= A) && (G <= C) && (H >= D)) {                       //overlapping upper left corner
area3 = (G - A)*(D - F);
}
else if ((F <= B) && (E <= A) && (G <= C) && (H <= D)) {                       //overlapping bottom left corner
area3 = (G - A)*(H - B);
}
else if ((F <= B) && (E >= A) && (G >= C) && (H <= D)) {                        //overlapping bottom right corner
area3 = (H - B)*(C - E);
}
else if (((C - A) <= (G - E)) && (H <= D) && (G >= C) && (E <= A) && (F <= B)) {               //overlapping bottom side
area3 = (C - A)*(H - B);
}
else if (((C - A) <= (G - E)) && (H >= D) && (G >= C) && (E <= A) && (F >= B)) {               //overlapping top side
area3 = (C - A)*(D - F);
}
else if (((D - B) <= (H - F)) && (E <= A) && (F <= B) && (H >= D) && (G <= C)) {               //overlapping left side
area3 = (G - A)*(D - B);
}
else if (((D - B) <= (H - F)) && (E >= A) && (F <= B) && (H >= D) && (G >= C)) {               //overlapping right side
area3 = (C - E)*(D - B);
}
else if (((C - A) >= (G - E)) && (E >= A) && (F >= B) && (C >= G) && (D <= H)) {      //overlapping part of top side
area3 = (G - E)*(D - F);
}
else if (((C - A) >= (G - E)) && (A <= E) && (B >= F) && (G <= C) && (D >= H)) {       //overlapping part of bottom side
area3 = (G - E)*(H - B);
}
else if (((D - B) >= (H - F)) && (E <= A) && (F >= B) && (G <= C) && (H <= D)) {      //overlapping part of left side
area3 = (G - A)*(H - F);
}
else if (((D - B) >= (H - F)) && (E >= A) && (F >= B) && (G >= C) && (H <= D)) {       //overlapping part of right side
area3 = (C - E)*(H - F);
}
else if (((G - E) <= (C - A)) && (E >= A) && (F <= B) && (G <= C) && (H >= D)) {     //overlapping top and bottom
area3 = (G - E)*(D - B);
}
else if (((H - F) <= (D - B)) && (E <= A) && (F >= B) && (C <= G) && (D >= H)) {     //overlapping left and right
area3 = (C - A)*(H - F);
}

return (area1 + area2 - area3);
}