d258 223

"223-rectangle Area (rectangular region)" " leetcode-interview algorithm classic-java Implementation" "All topic Directory Index" code Download "Https://github.com/Wang-Jun-Chao" Original title Find the total area covered by two rectilinear rectangles in a 2D plane. Each rectangle is defined to bottom left corner and top right corner as shown in the figure.Rectangle AreaAssume that "total" is never beyond the maximum possible value of int.The mai

/** 223. Rectangle areas * 12.12 by Mingyang * Large area minus small area * Here we test instructions do not understand wrong oh, here is to find all the place is cover, not only by the two overlap cover place * so we need To find two of the sum of the area minus the middle repeat two times the small area*/ Public intComputearea (intAintBintCintDintEintFintG,intH) {intMinu = 0; if(E >= C | | G B) Minu= 0; ElseMinu= (Math.min (G, C)-Math.max (A, E

no.223 Rectangle AreaFind 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.Find the area where two rectangles intersectVertical rectangles, each of which is represented by a coordinate of two points in the lower left and upper right cornersIn fact, the intersecti

Tags: Post get nbsp path pre character false input git commandWhen adding a file to be submitted using git Add, if the file name is in Chinese, it will display garbled characters such as 274\232\350\256\256\346\200\273\347\273\223.Solution: At the bash prompt, enter:FalseIf the Core.quotepath is set to False, the characters above 0x80 will not be quote. Chinese display is normal.Reference:http://blog.csdn.net/tyro_java/article/details/53439537Linux gi

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.Problem Solving Ideas:The Java implementation is as follows: public int Computearea (int A, int B, int C, int D, int E, int F, int G, int H) { i

Title: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.Tips:This question to determine whether there are overlapping parts.Code:classSolution { Public: intComputearea (intAintBintCintDintEintFintGintH) {if(C B)return(c-a) * (d-b) + (G-E) * (H-F); return(c-a) * (d-b) + (G-E) * (h-f)-(min (c,g)-Max (A,e))

-E) + (d-b) * (C-A); We could also see this, since A is the left edge of C, E was the left edge of g,the left coordinate forThe overlaied rectangles is:---------------------------------------------------intleft =Math.max (A, E); the right coordinate forThe overlaied rectangles is:-------------------------------------------------------intright =math.min (C, G); The same analysis could apply to the vertical direction. Skill:to compute the overall area, we could use:total area-Overlaid area.Solutio

{2 Public:3 intComputearea (intAintBintCintDintEintFintGintH) {4 intoverlap=0;5 //Case 1:6 if(C>=ecD) {7Overlap = (C-E) * (H-B);8 }9 //Case 2:Ten if(C>=ecd) { OneOverlap = (C-E) * (dB); A } - //Case 3: - if(C>=ecD) { theOverlap = (C-E) * (dF); - } - //Case 4: - if(E>=aeF) { +Overlap = (C-E) * (H-F); - } + //Case 5: A if(A>=eaH) { atOverlap = (c-a) * (H-B); - } -

Topic: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.Answer:There's nothing to say, two rectangles added, minus the area of the coincident part if there is a coincident part1 classSolution {2 Public:3 intComputearea (intAintBintCintDintEintFintGintH) {4 intsum;5sum= (c-a) * (d-b) + (G-E) * (H

Topic: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.First determine whether two rectangles intersect, do not intersect directly return two rectangular area, the intersection needs to calculate the intersection of the lower left and right points of the coordinates, with two rectangular area minus the inte

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.Idea: Need to calculate the area of the square cover, the key is to calculate the area of the coincident area, using the following calculation method;On the left side of the coincident area (bottom) (right, top)left = Math.max (A, E);Bottom = Math.max (B, F);

leftmost part of the overlapping section is the leftmost side of the Rect1 (A)The rightmost part of the overlap may be the rightmost (C) of the rect1, or it may be the rightmost (G) of the Rect22, calculate the length(1) When D The topmost part of the overlapping section is the top of the Rect1 (D)The bottom of the overlapping section may be the bottom (B) of the rect1, or it may be the bottom of the Rect2 (F)(2) When D > H,The topmost part of the overlapping section is the top of the Rect2 (H)

Topic linksTitle Requirements: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.To better illustrate the problem, here are a few diagrams specifically drawn:　　The area we require = two rectangular area

1. Description of the problemFind the area of two rectangles.　　2. Methods and IdeasThe intersection area of two rectangles can be calculated first, and then the area is rectangular with two rectangular area and minus intersection area.Note: Although the area does not exceed the maximum value of int, the middle edge length operation may be more than, pay attention to the processing details, otherwise easily overflow.　　Class Solution {public:intComputearea (intAintBintCintDintEintFintGintH) {intUn

Rectangle AreaTotal Accepted: 2205 Total Submissions: 8138 Find the total area covered by and rectilinear rectangles in a2D 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.IdeasThe area of two regions is calculated, and then the area of the overlapping is subtracted, which is the desired one.[CODE]public class Solution {public int computearea (int A, int B, int C, in

\) /c1> with \ (j-i+1-1\), they is the same!So our tasks now are to calculate \ (fac_{i,1}^{k}\), which is \ (i + k-2 \choose i-1\). But we cannot pre process the factorial of \ (k\), as \ (k\) can is up to \ (10^9\). But \ (fac_{1,1}\) are always 1!Think the transfer between \ (fac_{i-1,1}\) and \ (fac_{i,1}\), that is:\[\frac{(i-1+k-2)!} {(I-1-1)! (K-1)!} \rightarrow \frac{(i+k-2)!} {(I-1)! (K-1)!} \]We just need to multiply the first expression by \ (i+k-2\) and \ ((i-1) ^{-1}\, (mod\,10^9+7)

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