Leetcode 223: Rectangle Area, leetcoderectangleRectangle AreaTotal Accepted:2205Total Submissions:8138
Find the total area covered by twoRectilinearRectangles in2DPlane.
Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.
Assume that the total area is never beyond the maximum possible valueInt.
[Idea]
Obtain the area of the two regions, and then subtract the area of the overlapping.
[CODE]
public class Sol
Leetcode 223 Rectangle Area, leetcoderectangle1. Problem Description
Calculate the area of the two rectangles. 2. methods and ideas
You can first obtain the intersection area of the two rectangles, and then use the sum of the two rectangles minus the intersection area to represent the rectangle and area.Note: although the area does not exceed the maximum value of int, the length of the middle side may exceed. Pay attention to the handling details; ot
Original question:223. Little Kings
Time limit per test:0.5 sec.Memory limit per test:65536 KBInput:standardOutput:standard
After solving nice problems on bishops and rooks, Petya decided that he would like to learn to play chess. He started to learn, the rules and found out, the most important piece in the game is the king.
The king can move to any adjacent cell (there is up to eight such cells). Thus, Kings is in the attacking position, if they is l
the above (1) and (2):(1) TextView TV = new TextView (mainactivity.this); TV is dependent on activity (the interface exists), activity is destroyed, and TV is destroyed .If you use TextView TV = new TextView (getapplicationcontext ()), the activity may be destroyed, But the entire application has not been destroyed, so the TV will become a null pointer, causing memory leaks. (2) Alertdialog.builder builder = new Builder (mainactivity.this);The same dialog box is created, and the dialog box is a
Find the total area covered by and rectilinear rectangles in a 2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.For this problem, if the idea is correct, the solution is relatively simple.1. If there is no overlap, the total area of the two rectangles is returned directly: (C-A) * (d-b) + (G-E) * (h-f). Determine if two rectangles overlap: if (C2. If there is overlap,
Find the total area covered by and rectilinear rectangles in a 2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.public class Solution {public int Computearea (int A, int B, int C, int D, int E, int F, int G, int H) {int left = Math.max (A, E);int right = Math.min (C, G);int bottom = Math.max (B, F);int top = Math.min (D, H);int shared = Math.max (right-left, 0) * Math.
Find the total area covered by and rectilinear rectangles in a 2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.Credits:Special thanks to @mithmatt for adding this problem, creating the above image and all test cases.Subscribe to see which companies asked this question1 Public classSolution {2 Public intComputearea (intAintBintCintDintEintFintGintH) {3 in
, 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 "" " ifB>=horE>=corF>=dorA>=g:return(c-a) * (d-b) + (G-E) * (h-f) width = min (c,g)-Max (a,e) height = min (d,h)-Max (B,F)return(c-a) * (d-b) + (G-E) * (h-f)-Width * heightIdea twoCalculates the area of overlapping portions directly, with zero length or width when not overlapped.Code Oneclass Solution(object): def computeArea(self, A, B, C, D, E,
LeetCode 223 Rectangle Area (rectangular Area)Translation
Find the total area of the two intersecting rectangles in the two-dimensional plane. Each rectangle defines the coordinates in the lower left corner and the upper right corner. (For example, a rectangle) assumes that the total floor space will never exceed the maximum value of an int.
Original
Analysis
I tried this question the day before yesterday and wrote a bunch of judgments. After all, it
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