leetcode-223-Rectangle-Area

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描述

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.
rec

分析

最重要的的是找到两个长方形的相交面积的表达式，画个草图就理解了，所以说，编程还是得数学好。需要注意的一点是，在判断中，应该直接做判断，比如if(l2 > l1 && h2 > h1),而不是if(l2-l1 > 0 && h2-h1>0)，因为这样有可能溢出（这个bug让我折腾好久），这个边界情况应该注意，今后写代码也是如此。

解决方案1(C++)

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class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int l1 = max(A, E);
        int l2 = min(C, G);
        
        int h1 = max(B, F);
        int h2 = min(D, H);
        int extra_area = 0;
        
        if(l2 > l1 && h2 > h1) {
            extra_area = (l2-l1) * (h2-h1);    
        }else {
            extra_area = 0;
        }
        
        return (G-E)*(H-F)+(C-A)*(D-B)-extra_area;
    }
};

解决方案2(Java)

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public class Solution {
    public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
    	
    	int l1 = Math.max(A, E);
        int l2 = Math.min(C, G);
        
        int h1 = Math.max(B, F);
        int h2 = Math.min(D, H);
        int extra_area = 0;
        
        if(l2 > l1 && h2 > h1) {
            extra_area = (l2-l1)*(h2-h1);    
        }else {
            extra_area = 0;
        }
        
        return (G-E)*(H-F)+(C-A)*(D-B)-extra_area;
    }        
}

解决方案3(Python)

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class Solution(object):
    def computeArea(self, A, B, C, D, E, F, G, H):
        """
        :type A: int
        :type B: int
        :type C: int
        :type D: int
        :type E: int
        :type F: int
        :type G: int
        :type H: int
        :rtype: int
        """
        l1 = max(A, E)
        l2 = min(C, G)
        
        h1 = max(B, F)
        h2 = min(D, H)
        extra_area = 0
        
        if(l2 > l1 and h2 > h1):
            extra_area = (l2-l1) * (h2-h1)
        else:
            extra_area = 0
        
        return (G-E)*(H-F)+(C-A)*(D-B)-extra_area

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