Leetcode:trapping Rain Water II

The height of each unit cell was greater than 0 and is less than 20,000. Example:given the following 3x6 height map:[  [1,4,3,1,3,2],  [3,2,1,3,2,4],  [ 2,3,3,2,3,14.

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, water is trapped between the blocks. The total volume of water trapped is 4.

Refer to HTTPS://DISCUSS.LEETCODE.COM/TOPIC/60418/JAVA-SOLUTION-USING-PRIORITYQUEUE/2

Here's a concrete example: http://www.cnblogs.com/grandyang/p/5928987.html

Analysis, according to the principle of the cask, first find the periphery of the shortest bar, inside if there is a bar than it is shorter, must be able to save water (because four weeks all bar is higher than it)

Note that it is possible to save more water, as it is likely that the cell will change in height. So to put the high bar in the middle of the BFS into the queue, as the water level rises to the height of these bars to see if there is a groove to save more

1  Public classSolution {2      Public classCell {3         intRow;4         intCol;5         intheight;6          PublicCell (intXintYintval) {7              This. Row =x;8              This. Col =y;9              This. Height =Val;Ten         } One     } A      -      Public intTraprainwater (int[] heightmap) { -         if(heightmap==NULL|| heightmap.length<=2 | | heightmap[0].length<=2)return0; the         intm =heightmap.length; -         intn = heightmap[0].length; -         intres = 0; -priorityqueue<cell> queue =NewPriorityqueue<cell> (1,NewComparator<cell>() { +              Public intCompare (cell C1, cell C2) { -                 returnC1.height-C2.height; +             } A         }); atHashset<integer> visited =NewHashset<integer>(); -          for(inti=0; i<m; i++) { -Queue.offer (NewCell (i, 0, heightmap[i][0])); -Queue.offer (NewCell (i, n-1, heightmap[i][n-1])); -Visited.add (i*n+0); -Visited.add (i*n+n-1); in         } -          for(intj=0; j<n; J + +) { toQueue.offer (NewCell (0, J, heightmap[0][j])); +Queue.offer (NewCell (M-1, J, Heightmap[m-1][j])); -Visited.add (0*n+j); theVisited.add ((m-1) *n+j); *         } $         int[] Directions =New int[][]{{-1, 0}, {1, 0}, {0, 1}, {0,-1}};Panax Notoginseng          while(!Queue.isempty ()) { -Cell cur =Queue.poll (); the              for(int[] dir:directions) { +                 introw = Cur.row + dir[0]; A                 intCol = Cur.col + dir[1]; the                 if(row>=0 && row<m && col>=0 && col<n &&!visited.contains (row*n+col)) { +Visited.add (row*n+col); -Res + = Math.max (0, Cur.height-Heightmap[row][col]); $Queue.offer (NewCell (Row, col, Math.max (cur.height, Heightmap[row][col ))); $                 } -             } -         } the         returnRes; -     }Wuyi}

Leetcode:trapping Rain Water II