trapping for beginners

Trapping Rain WaterGiven n non-negative integers representing an elevation map where the width of each bar are 1, compute how much water it is able to trap after raining.For example,Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of the Rain Water (blue section) is being trapped. Thanks Marcos for contributing this image!Idea: This question is easy to see, but t

Given m x n a matrix of positive integers representing the height of each unit cell in a 2D elevation map, compute the VO Lume of water it is able to trap after raining.Note:Both m and N is less than 110. The height of each unit cell is greater than 0 and was less than 20,000.Example:Given the following 3x6 height map:[ [1,4,3,1,3,2], [3,2,1,3,2,4], [2,3,3,2,3,1]]return 4.The above image represents the elevation map [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] before the rain.After the rain, w

Https://leetcode.com/problems/trapping-rain-water/#/solutionsHere's my idea:instead of calculating area by Height*width, we can think it in a cumulative. In the other words, the sum water amount of each bin (width=1).Search from left to right and maintain a max height of left and right separately, which was like a one-side wall of partial Container. Fix the higher one and flow water from the lower part. For example, if current height of lower, we fill

https://leetcode.com/articles/trapping-rain-water/Finish a hard difficulty of the topic, the individual is very excited, suddenly that feel "double pointer" is simply an artifact, the title requirements of the entire container can be filled with the amount of water, intuitively, only the groove part can be filled with water, violent search method is too high complexity, the best way is to use a double pointer, left and right to place one, Notice how m

Given n non-negative integers representing an elevation map where the width of each bar are 1, compute how much WA ter It is the able to trap after raining.For example,Given [0,1,0,2,1,0,1,3,2,1,2,1] , return 6 .The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1].In this case, 6 units of the Rain Water (blue section) is being trapped. Thanks Marcos for contributing this image!classSolution { Public: intTrapintA[],intN) {if(a==null| | n3) return 0; intmaxleft=0; in

],right_most[i])-A[i]) - returnSumIdeas :Bottom line: How much water a position can put on, depending on the smallest of the left and right sides with this position position high.Imagine the physical environment in which a location can hold water, ensuring that the position is in a low-lying position. How can we meet the conditions of the low-lying position? Both sides have to have a higher element than this position. What if we can guarantee that there are more elements on both left and

Given n non-negative integers representing an elevation map where the width of each bar are 1, compute how much WA ter It is the able to trap after raining.For example,Given [0,1,0,2,1,0,1,3,2,1,2,1] , return 6 .Reference:http://blog.csdn.net/doc_sgl/article/details/123071711 Public classSolution {2 Public intTrapint[] A) {3 intWatersum = 0;4 intLeftmaxheight[] =New int[a.length];5 intRightmaxheight[] =New int[a.length];6 7 intMaxHeight = 0;8

continue to sweep, and so on, this method can solve the problem above, but The last interval needs to be considered separately .In fact, this can be solved: first find the highest problem maxhigh, and then separate from both sides to this node traversal, set a secondary high Temphigh,If the current node is higher than the secondary high, then update the secondary highs, continue;Otherwise, the current node is subtracted from the secondary highs, which is the volume of water we can hold.It is s

there is a groove on the line (large, middle small), not necessarily to find a value larger than the current value, such as the sequence [4,2,3], then when N (not found larger than num[i), take min (Num[n-1],num[i]) as the river Height, (j-i-1) as the river width, minus the middle of the reef (temp) can it? Obviously not, if the sequence is [4,2,3,1] This does not work, and finally I came up with the method is to cut the dike, that is, when unable to find a larger value than num[i], I will redu

Given n non-negative integers representing an elevation map where the width of each bar are 1, compute how much WA ter It is the able to trap after raining.For example,Given [0,1,0,2,1,0,1,3,2,1,2,1] , return 6 .I think this problem is similar to the question of buying stocks and the maximum water storage, and I feel that I should do it in a divide-and-conquer way.But not ... This is the answer given by someone else.The idea is too ingenious ... Ah ....An O (n) solution are to consider each bar

No brain Code Code (TLE):1#include 2#include 3 4 using namespacestd;5 6 intFindleft (intI, vectorint>height)7 {8 intt = I1;9 while((t+10)Ten { One if(Height[t] > height[t +1]) At--; - Else - Break; the } - returnT +1; - } - + intFindright (intI, vectorint>height) - { + intt = i +1; A while((t1>=0) t height.size ()) at { - if(Height[t] > Height[t-1]) -t++; - Else - Break; - } in returnT-1; - } to

[LeetCode] [Java] Trapping Rain WaterQuestion: Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. For example,Given[0,1,0,2,1,0,1,3,2,1,2,1], Return6. The above elevation map is represented by array [,]. in this case, 6 units of rain water (blue section) are being trapped. thanks Marcos for contributing this image!Question: Given n non-negative intege

Article title: trapping intruders in Linux (2 ). Linux is a technology channel of the IT lab in China. Includes basic categories such as desktop applications, Linux system management, kernel research, embedded systems, and open source. Author: Hong Xiaoye 　　 Install 　　 It should be noted that installing a host-based intrusion detection system does not provide protection for the system. It aims to help administrators analyze and record potential attack

Article title: trapping intruders in Linux (3 ). Linux is a technology channel of the IT lab in China. Includes basic categories such as desktop applications, Linux system management, kernel research, embedded systems, and open source. Author: Hong Xiaoye 　　 Configuration 　　 After installing SNARE and running it, you need to configure it. Just as the syslogd monitoring program has syslog. conf, the auditd monitoring program also has the audit. conf fi

(int First,int second) { if (first>100| | SECOND>100) Throw new Iamsecondgrade (); Throw Exception return (First*second); } Public static void Main () { int mul_value; Try { Mul_value=mul (99,56); System.Console.WriteLine ("99 and 56 product: {0}", Mul_value); Mul_value=mul (101,4); System.Console.WriteLine ("There is an exception, this line will not be executed.")"); } catch (Iamsecondgrade)//Capture Custom exceptions { System.Console.WriteLine

(), Data:eval ("("+ $("#data"). Val () +")"), success:function (results) {if(typeofResults = ="Object") { $("#result"). HTML (json.stringify (results)); } Else { varResultobj =Json.parse (results); $("#result"). HTML (results); } //when used, it needs to be converted to a JSON object}, Error:function (XMLHttpRequest, Textstatus, Errorthrown) {alert (Xmlhttprequest.responsetext); }

same time update waterTen for(inti =0; i ) One if(A[i] >left_max_height) ALeft_max_height =A[i]; - Else -Water + = left_max_height-A[i]; the intRight_max_height =0; - //Calc The right_max_height, at the same time update water - for(inti = n-1; i > Max; i--) - if(A[i] >right_max_height) +Right_max_height =A[i]; - Else +Water + = right_max_height-A[i]; A returnwater; at

Container with most waterGivenNnon-negative integersa1 ,a2 , ...,an , where each represents a point at coordinate (I,ai ).NVertical lines is drawn such that the both endpoints of lineIis at (I,ai ) and (I, 0). Find lines, which together with X-axis forms a container, such that the container contains the most water.Note:you may not slant the container.Analysis reference here, the author explains it in detail.Class Solution {public: int. maxarea (vectorTrapping Rain Watergivenn non-negative in

Given n non-negative integers representing an elevation map where the width of each bar are 1, compute how much WA ter It is the able to trap after raining.For example,Given [0,1,0,2,1,0,1,3,2,1,2,1] , return 6 .The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1].In this case, 6 units of the Rain Water (blue section) is being trapped. Thanks Marcos for contributing this image!Reference: Dai Fangqin ([email protected]) Https://github.com/soulmachine/leetcodepublic class Sol

, and the last highest max value is the result.1 classSolution {2 Public:3 intTrap (vectorint>height) {4 if(height.size () = =0)return 0;5 intleft =0, right = Height.size ()-1, res =0;6 intMaxleft = Height[left], maxright =Height[right];7 while(Left Right )8 {9 if(Height[left] Height[right])Ten { One if(Height[left] > maxleft) maxleft =Height[left]; A Else if(Maxleft-height[left] >0) -res =Max (