[Leetcode] 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!

Method One:

Time complexity O (n), spatial complexity O (n)

For each pillar, find the tallest pillar on the left and right side, the area that the pillar can hold is min (leftmostheight,rightmostheight)-height. So

1. Scan from left to right, and for each pillar, the left maximum value is obtained;

2. Scan from right to left, for each column, the maximum right value;

3. Scan again and add up the area of each pillar.

Code

1 classSolution {2      Public: 3         intTrapintA[],intN)4         {5vector<int>Left ;6vector<int>Right ;7 left.resize (n);8 right.resize (n);9 Ten             //PrintArray (A, n); One  Aleft[0] ==0; -right[n-1] ==0; -              the             //get the left ' s Max -              for(inti =1; i< n;i++) -             { -Left[i] = max (left[i-1], a[i-1]); +             } -             //Printvector (left); +  A             //get the right ' s Max at              for(inti = n2; i>=0; i--) -             { -Right[i] = max (right[i+1], a[i+1]); -             } -             //Printvector (right); -  in             //Clac the Trap water -             intsum =0; to             intHeight =0; +              for(inti =0; i< n;i++) -             { theHeight =min (left[i], right[i]); *                 if(Height >A[i]) $Sum + = height-A[i];Panax Notoginseng             } -             returnsum; the         } +};

Method Two:

Time complexity O (n), Spatial complexity O (1)

1. Scan once and find the highest pillar, which divides the array into two halves;

2. Handle the left half;

3. Handle the right half.

1 classSolution {2      Public:3         intTrapintA[],intN) {4             intMax =0; 5              for(inti =0; I < n; i++)6                 if(A[i] > A[max]) max =i;7             intWater =0;8             intLeft_max_height =0;9             //Calc The left_max_height, at the same time update waterTen              for(inti =0; i < Max; 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         }    -};

Method 3:

// LeetCode, Trapping Rain Water // 用一个栈辅助，小于栈顶的元素压入，大于等于栈顶就把栈里所有小于或 // 等于当前值的元素全部出栈处理掉，计算面积，最后把当前元素入栈 // 时间复杂度 O(n)，空间复杂度 O(n)can be used std::p air insteadstruct node structure

1 structNode2 {3     intVal;4     intindex;5 Node () {}6Node (intVintidx): Val (v), index (IDX) {}7 };8 9 classSolution {Ten      Public: One         intTrapintA[],intN) { AStack<node>s; -             intsum =0; -              for(inti =0; I < n; i++) the             { -                 intHeight =0; -         //dispose of elements in the stack that are shorter or higher than the current element -                  while(!s.empty ()) +                 { -Node node =s.top (); +             //If you encounter a higher than the current element, calculate the area first, such as {4,3,2}, and then jump out of the while loop, pressing the current element stack A                     if(A[i] < Node.val)//a[] = {4, 2,3}, Calc sum at                     { -                         intwidth = i-node.index-1; -Sum + = a[i] * Width-height *width; -                          Break; -                     } -                     Else in                     { -             //node.val, height, a[i] The hollow of the three ; to                         intwidth = i-node.index-1; +Sum + = Node.val * Width-height *width; -Height = node.val;//Update Height the                 //pop up the top of the stack because the element is finished and no longer needed * S.pop (); $                     }Panax Notoginseng                 } -         //all elements must be in the stack the S.push (Node (a[i], i)); +             } A  the             returnsum; +         } -};