【題目】

題目連結
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.

【分析】

簡單題：
注意點就是，判斷語句：
if (E > C || G < A || F > D || H < B)不要用min,max然後相減去判斷，不然可能因為相減得到的暫時結果越界
比如

public class Solution {  
    public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {  
        int area1 = (C-A) * (D-B);  
        int area2 = (G-E) * (H-F);  

        int overlapRegion = overlap(A, B, C, D, E, F, G, H);  
        return area1 + area2 - overlapRegion;  
    }  

    private
 
 int overlap(int A, int B, int C, int D, int E, int F, int G, int H) {  
        int h1 = Math.max(A, E);  
        int h2 = Math.min(C, G);  
        int h = h2 - h1;  

        int v1 = Math.max(B, F);  
        int v2 = Math.min(D, H);  
        int v = v2 - v1;  

        if(h<=0 || v<=0) return
 
 0;  
        else return h*v;  
    }  
}  

其中的h,v就可能越界
所以要這樣寫：

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