Surrounded Regions--leetcode

Given a 2D board containing ‘X‘ ‘O‘ and, capture all regions surrounded by ‘X‘ .

A region was captured by flipping all ‘O‘ s into ‘X‘ s-surrounded region.

For example,

x x x xx o o xx x o xx o x x

After running your function, the board should is:

x x x xx x x xx x x xx O x x

Title: Give a chessboard with ' X ' and ' O ', find all ' o ' surrounded by ' x ' and replace it with ' X '.

Problem Solving Ideas:

Solution one: From the edge, find out on the four side o, in the way of BFS to find all o connected to the Edge O, these o are not surrounded, the O is replaced with a special character, after traversing, the O is replaced by x, the special character is replaced with O.

Solution two: Directly with the BFS traversal, find an O, join the BFS queue, find out all the O connected with it at the same time to determine whether the O is surrounded (that is, whether on a certain edge), and add these o to a list, Each connected O-Zone uses a surround Boolean variable to indicate whether the connected area is surround, and if it is replaced with X, it is not replaced and continues to traverse the other.

Note: Here because I, j subscript into the queue or result set, I started with the string type key= i+ "_" +j to deal with, and later saw someone else key = i * Collen + j; Collen is the number of columns of the chessboard, and then uses Key/collen,key%collen to get the i,j subscript more efficient.

Talk is cheap>>

Solution One:

 Public voidSolveChar[] board) {Queue<Integer> queue =NewArraydeque<>(); intRowlen =board.length; if(Rowlen = = 0)            return; intCollen = board[0].length; int[] adj = {{0,-1}, {0, 1}, {-1, 0}, {1, 0}};  for(inti = 0; i < Rowlen; i++) {             for(intj = 0; J < Collen; J + +) {
If O is on four sidesif(Board[i][j] = = ' O ' && (i = = 0 | | j = 0 | | i = = RowLen-1 | | j = = ColLen-1)) {                    intKey = i * Collen +J; BOARD[I][J]= ' 1 ';                    Queue.add (key);  while(!Queue.isempty ()) {
BFS traversal O-Key with O-connected on edge=Queue.poll (); intx = key/Collen; inty = key%Collen;  for(intk = 0; K < 4; k++) {                            intadj_x = x + adj[k][0]; intadj_y = y + adj[k][1]; if(adj_x >= 0 && adj_y >= 0 && adj_x < rowlen && Adj_y <Collen) {                                if(board[adj_x][adj_y] = = ' O ') {                                    intpos = adj_x * Collen +adj_y; Board[adj_x][adj_y]= ' 1 ';                                Queue.add (POS);        }                            }                        }                    }                }            }        }  for(inti = 0; i < Rowlen; i++) {             for(intj = 0; J < Collen; J + +) {                if(Board[i][j] = = ' O ') {Board[i][j]= ' X '; } Else if(Board[i][j] = = ' 1 ') {Board[i][j]= ' O '; }            }        }    }

Solution Two:

 Public voidSolveChar[] board) {Queue<Integer> queue =NewArraydeque<>(); intRowlen =board.length; if(Rowlen = = 0) {            return; }        int[] adj = {{0,-1}, {0, 1}, {-1, 0}, {1, 0}}; intCollen = board[0].length; Boolean[] visited =New Boolean[Rowlen][collen];  for(inti = 0; i < Rowlen; i++) {             for(intj = 0; J < Collen; J + +) {                if(!visited[i][j] && board[i][j] = = ' O ') {
Standard BFS TraversalBooleanSurround =true; List<Integer> Tohandle =NewArraylist<>(); Queue.add (i* Collen +j);  while(!Queue.isempty ()) {                        intKey =Queue.poll ();                        Tohandle.add (key); intx = key/Collen; inty = key%Collen;  for(intk = 0; K < 4; k++) {
Check o up or downintadj_x = x + adj[k][0]; intadj_y = y + adj[k][1]; if(adj_x >= 0 && adj_y >= 0 && adj_x < rowlen && Adj_y <Collen) {
is not on the edge.if(board[adj_x][adj_y] = = ' O ' &&!Visited[adj_x][adj_y]) {                                    intpos = adj_x * Collen +adj_y;                                    Queue.add (POS); Visited[adj_x][adj_y]=true; }                            } Else {
There is a surround=false surround on the edge .=false; }                        }                    }                    if(surround) { for(intkey:tohandle) {Board[key/collen][key% Collen] = ' X '; }                    }                }            }        }    }

Surrounded Regions--leetcode