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  • 223. 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.

    Rectangle Area

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

    链接: http://leetcode.com/problems/rectangle-area/

    题解:

    数学题,需要判断矩阵是否相交,相交的话减去重复面积(顶点相交除外)。

    Time Complexity - O(1), Space Complexity - O(1)。

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

    二刷:

    方法跟一刷一样

    Java:

    Time Complexity - O(1), Space Complexity - O(1)。

    public class Solution {
        public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
            int totalArea = (C - A) * (D - B) + (G - E) * (H - F);
            if (A >= G || B >= H || C <= E || D <= F) {
                return totalArea;
            }
            int sameArea = (Math.min(C, G) - Math.max(A, E)) * (Math.min(D, H) - Math.max(B, F));
            return totalArea - sameArea;
        }
    }

    三刷:

    Java:

    1. 一开始先计算出两个矩形的面积和 totalArea
    2. 判断两个矩形是否相交,假如不相交,或者仅有顶点相交,那么我们直接返回totalArea。 这里两个矩形 x 的范围是 (A, C), (E, F),  y的范围是(B, D), (E, F)
    3. 计算overlap的面积,边的计算公式是 ( 最小的上方或者右方点 -  最大的下方或者左方点), 相乘就是overlap的面积
    4. 相减得到结果
    public class Solution {
        public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
            int totalArea = (C - A) * (D - B) + (G - E) * (H - F);
            if (A >= G || C <= E || B >= H || D <= F) {
                return totalArea;
            }
            int overlap = (Math.min(C, G) - Math.max(A, E)) * (Math.min(D, H) - Math.max(B, F));
            return totalArea - overlap;
        }
    }

    Reference:

    https://leetcode.com/discuss/39188/an-easy-to-understand-solution-in-java

    https://leetcode.com/discuss/39398/my-java-solution-sum-of-areas-overlapped-area

    https://leetcode.com/discuss/43549/just-another-short-way

    https://leetcode.com/discuss/43173/if-you-want-to-laugh-look-at-my-solution

    https://leetcode.com/discuss/54138/python-concise-solution

    https://leetcode.com/discuss/51354/an-explanation-in-plain-language

    http://www.cnblogs.com/0001/archive/2010/05/04/1726905.html

    http://www.geeksforgeeks.org/find-two-rectangles-overlap/

    https://www.cs.princeton.edu/~rs/AlgsDS07/17GeometricSearch.pdf

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  • 原文地址:https://www.cnblogs.com/yrbbest/p/4993512.html
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