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223. 矩形面积

English Version

题目描述

在二维平面上计算出两个由直线构成的矩形重叠后形成的总面积。

每个矩形由其左下顶点和右上顶点坐标表示，如图所示。

示例:

输入: -3, 0, 3, 4, 0, -1, 9, 2
输出: 45

说明: 假设矩形面积不会超出 int 的范围。

解法

计算重叠部分的面积，注意考虑没有重叠的情况

Python3

class Solution:
    def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
        a = (ax2 - ax1) * (ay2 - ay1)
        b = (bx2 - bx1) * (by2 - by1)
        width = min(ax2, bx2) - max(ax1, bx1)
        height = min(ay2, by2) - max(ay1, by1)
        return a + b - max(height, 0) * max(width, 0)

Java

class Solution {
    public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        int a = (ax2 - ax1) * (ay2 - ay1);
        int b = (bx2 - bx1) * (by2 - by1);
        int width = Math.min(ax2, bx2) - Math.max(ax1, bx1);
        int height = Math.min(ay2, by2) - Math.max(ay1, by1);
        return a + b - Math.max(height, 0) * Math.max(width, 0);
    }
}

C++

class Solution {
public:
    int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        int a = (ax2 - ax1) * (ay2 - ay1);
        int b = (bx2 - bx1) * (by2 - by1);
        int width = min(ax2, bx2) - max(ax1, bx1);
        int height = min(ay2, by2) - max(ay1, by1);
        return a + b - max(height, 0) * max(width, 0);
    }
};

Go

func computeArea(ax1 int, ay1 int, ax2 int, ay2 int, bx1 int, by1 int, bx2 int, by2 int) int {
	a := (ax2 - ax1) * (ay2 - ay1)
	b := (bx2 - bx1) * (by2 - by1)
	width := min(ax2, bx2) - max(ax1, bx1)
	height := min(ay2, by2) - max(ay1, by1)
	return a + b - max(height, 0)*max(width, 0)
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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