trapping for beginners

columns that are not allowed to be duplicated, or by uniquely setting a unique value attribute for a single column.Statistical data duplication: Statistics in the person table LastName, FirstName are not repeated and the number of records more than 1 dataSelect COUNT (*) as rep, lastname,firstname from personGROUP BY Lastname,firstnameHave on rep>1;Mode of thinking:Determine the repeating column LastName FirstNameLists the repeating column and count (*) in the column selection list (the field a

intCuridx =0;Ten Onestackint>Stk; A - intRetsum =0; - for(; curidx){ the if(Stk.empty ()) { - Stk.push (CURIDX); - Continue; - } + intStktop =stk.top (); - if(a[stktop]>=A[curidx]) { + Stk.push (CURIDX); A Continue; at } - while(!Stk.empty ()) { - intDIT =Stktop; - Stk.pop (); - if(Stk.empty ()) Break; -Stktop =stk.top (); inRetsum +

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!Search from both sides to the middle. First find the maximum value and then the head to the maximum

Trapping Rain WaterProblem: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.Ideas:Array of special, get the highest point value, left and right side of the test separatelyMy Code: Public classSolution { Public intTrapint[] A) {if(A = =NULL|| A.length = = 0)return0; intMaxindex =-1; intMaxValue =Integer.min_value; for(inti = 0; i ) {

down (right-to-left) and sweep it all over again in the same way.But there is one more problem, look at the situation below. When the left and right borders (pointed by the Red Arrows) are high, the capacity of this part of the water will be calculated two times (sweep from left to right once, and one at the left).The solution is also very simple, as long as the left-to-right boundary must be less than the right boundary, and the left boundary must be greater than or equal to the right boundary

This article is Senlie original, reprint please retain this address:Http://blog.csdn.net/zhengsenlieTrapping Rain WaterTotal Accepted: 14568 Total Submissions: 50810 My SubmissionsGiven 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

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.Idea: The amount of water that a certain lattice can hold depends on the lower of the two tallest lattices on either side. Consider traversing from both sides toward the middle, where height represents the current storage watermark, and if A[left]>=height a[right]>=height, the height is s

https://leetcode.com/problems/trapping-rain-water/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!Prob

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!Class Solution {public: int trap (int a[], int n) { if (n Leetcode-

https://oj.leetcode.com/problems/trapping-rain-water/Simulate the problem and calculate the amount of water stored in the bump.For a position I, calculate the maximum left (scan from left), and the maximum right (scan from right). Find the minimum value in right, if it is greater than a[i], then do min-a[i] accumulation.classSolution { Public: intTrapintA[],intN) {if(n1) return 0; Vectorint>Lefthigher; Vectorint>Righthigher; intLeftlarge

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] , Return6 . 650) this.width=650; "src=" Http://www.leetcode.com/wp-content/uploads/2012/08/rainwatertrap.png "style=" border:0 Px;vertical-align:middle; "alt=" Rainwatertrap.png "/>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 (blu

Resolution Reference: Http://www.xuebuyuan.com/1586534.htmlbegan to be that to the figure of the cheat, thought as long as the drop up can be, and in fact, some of the middle of the highest point, and can not feel the whole rain height.Finally, two pointers are traversed in the middle, each moving smaller, and the current maximum is not moving. Calculate the area to use the current second-highest line.1 classSolution {2 Public:3 intMin (intAintb)4 {5 returnA>b?b:a;6 } One

finally add up is the total capacity.Code:#include #include#includeusing namespacestd;intMaxtrappingwater (Constvectorint> water) { intsz=water.size (); Vectorint>Premaxwater (SZ); premaxwater[0]=0; for(intI=1; i){ if(water[i]>premaxwater[i-1]) Premaxwater[i]=Water[i]; ElsePremaxwater[i]=premaxwater[i-1]; } Vectorint>Sufmaxwater (SZ); Sufmaxwater[sz-1]=0; for(inti=sz-2; i>=0; i--){ if(water[i]>sufmaxwater[i+1]) Sufmaxwater[i]=Water[i]; ElseSufmaxwater[i]=sufmaxwater[i+1]; }

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!Method One: Collect rainwaterOn the left, the short side is the dam. Shorter than a levee can store

Problem Description: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!Basic ideas:Through analysis, it can be found that the tank of the storage wat

Before tap, to catch exceptions to a task in a thread, there are usually two ways:1. Blocking: After the thread starts, at the appropriate time, call wait, or the WaitAll method.2, non-blocking (recommended): When the task is set up, write the ContinueWith method of the task and catch the exception in the method.For the tap era, if we were to capture a task with the Asyn keyword, the ContinueWith method would capture the failure. It is also not possible to register a global static method.Blockin

right side of the traversal method is similar. Such as:Runtime:8msClass Solution {public: int Trap (vectorSolution TwoThe second solution is to use a single traversal method.The above solution is to traverse two times, in order to find the maximum value, so as to ensure that the difference between the maximal value and the current value is the amount of water can be accommodated, but in fact can not find the maximum value but the left and right of the current value of the maximum value. If t

This problem has an intuitive idea, is to record the left and right of each point maximum value for height[i] This point can be installed water equals min (Leftmax[i], rightmax[i])-height[i]. The solution needs to be scanned 2 times in sequence.The following methods only need to scan the sequence once.classSolution:#@param {integer[]} height #@return {integer} defTrap (self, height): Stack=[] ans=0 I, length=0, len (height) whileI Length:ifstack = =[]: Stack.append ([Height[i],1])

Problem descriptionTrapping Rain WaterGiven 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!Problem analysis This topic solution thinking can refer to Le

straight down step, cycle directly end the Break; - } Wu intStepwater = 0; - intK = 0; About if(Union.size () > 0) {//Calculate Stairs $ while(K Height[lower]) { -Stepwater + = Height[i]-height[i + k + 1];//Detail 3, calculate the amount of water on step -k++; - } A } + //In addition, the Maxpoten should be cut off to lower, lower after th