LeetCode:矩形区域【223】

LeetCode:矩形区域【223】

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。

Rectangle Area

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。

题目分析

这道题目应该很简单,但是思路一定清晰,即搞清楚矩形重叠的条件

Java题解

public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        
        int areaOfSqrA = (C-A) * (D-B);
         int areaOfSqrB = (G-E) * (H-F);
        
        int left = Math.max(A, E);
        int right = Math.min(G, C);
        int bottom = Math.max(F, B);
        int top = Math.min(D, H);
        
        //If overlap
        int overlap = 0;
        if(right > left && top > bottom)
             overlap = (right - left) * (top - bottom);
        
        return areaOfSqrA + areaOfSqrB - overlap;
}

  

 

posted @ 2018-08-15 10:35  子烁爱学习  阅读(188)  评论(0编辑  收藏  举报