leetcode之矩形面积

• 题目
• 题解
• • 我的题解：
  • 官方题解：

题目

题解

我的题解：

public class m223 {
    public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        int visualArea = getArea(ax1, ay1, ax2, ay2) + getArea(bx1, by1, bx2, by2);
        if (ax2 < bx1 || ay1 > by2 || ax1 > bx2 || ay2 < by1) {
            return visualArea;
        }

        int overlaps = getArea(Math.max(ax1, bx1), Math.max(ay1, by1), Math.min(ax2, bx2), Math.min(ay2, by2));

        return visualArea - overlaps;

    }

    public int getArea(int x1, int y1, int x2, int y2) {
        return (x2 - x1) * (y2 - y1);
    }
}

官方题解：

class Solution {
    public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        int area1 = (ax2 - ax1) * (ay2 - ay1), area2 = (bx2 - bx1) * (by2 - by1);
        int overlapWidth = Math.min(ax2, bx2) - Math.max(ax1, bx1), overlapHeight = Math.min(ay2, by2) - Math.max(ay1, by1);
        int overlapArea = Math.max(overlapWidth, 0) * Math.max(overlapHeight, 0);
        return area1 + area2 - overlapArea;
    }
}

标签：ax2,ax1,bx2,int,面积,bx1,矩形,leetcode,Math
来源： https://blog.csdn.net/MoshMode/article/details/120589938