Rectangle Area

Find the total area covered by two rectilinear rectangles in a2D 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

[思路]

求出兩個區域的面積, 然後減去overlapping的區域, 即為所求.

[CODE]

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;
    }
}

leetcode 223: Rectangle Area