LeetCode 223. Rectangle Area

原题链接在这里：https://leetcode.com/problems/rectangle-area/

题目：

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.

Example:

Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2
Output: 45

Note:

Assume that the total area is never beyond the maximum possible value of int.

题解：

方块1面积 + 方块2面积 - 重叠面积

Note: 算重叠的面积时会有overflow, Math.min(C,G) 之前必须加 cast, e.g Math.min(C,G) = -150000001, Math.max(A,E) = 150000000.

原来写(long)(Math.min(C,G) - Math.max(A,E))会报错是因为Math.min(C,G)和Math.max(A,E)都是 Integer, 所以cast之前的结果会default成Integer, 还是会有overflow.

Time Complexity: O(1).

Space: O(1).

AC Java:

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);
        
        long width = Math.max((long)Math.min(C,G) - (long)Math.max(A,E), 0);
        long hight = Math.max((long)Math.min(D,H) - (long)Math.max(B,F), 0);
        int overflow = (int)(width*hight);
        
        return area1+area2-overflow;
    }
}

Could avoid overflow as well.

Time Complexity: O(1).

Space: O(1).

AC Java: 

class Solution {
    public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        return area(A, B, C, D) + area(E, F, G, H) - area(Math.max(A, E), Math.max(B, F), Math.min(C, G), Math.min(D, H));
    }
    
    private int area(int a, int b, int c, int d){
        if(a >= c || b >= d){
            return 0;
        }
        
        return (c - a) * (d - b); 
    }
}