[Leetcode] [Java] Surrounded regions

Source: Internet
Author: User

Title:

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`
Test Instructions:

Given a 2-dimensional planar inclusion `‘X‘` and `‘O‘,填充所有的被‘X‘包围的区域` .

Like what

`x x x xx o o xx x o xx o x x`
After you run your function, the plane becomes:

`x x x xx o o xx x o xx o x x`

Algorithm Analysis:

* Typical BFS topics. Traversing each character, if "O", starts the BFS traversal from the current character and joins the currently traversed queue if it is also surrounded by "o".

* until all adjacent "O" are traversed, at the same time, to determine whether each o is surrounded, as long as by an O is not surrounded by,

* The currently traversed set of O is not surrounded because these o are connected.

AC Code:

`<span style= "Font-family:microsoft yahei;font-size:12px;" >public class Solution {public void Solve (char[][] board) {if (board==null| |        board.length==0) return;        int row=board.length;        int col=board[0].length; Boolean visited[][] = new Boolean[row][col];//Traversal tag array for (int i=0;i<row;i++) {for (int J=0;j<co l;j++) {if (board[i][j]== ' O ' && (!visited[i][j])) {BFS (Board,i,j,vi sited);//If not traversed, then breadth-first search for the node}}}}private void BFs (char[][] board,int i,int J,boolean visited[]   []) {arraylist<integer> list =new arraylist<integer> ();        queue<integer> queue = new Linkedlist<integer> (); Boolean label=true;        int row=board.length;        int Col=board[0].length;int TEMJUDGE,TEMI,TEMJ; Queue.add (I*COL+J);//Convert the two-dimensional array position to one-dimensional visited[i][j]=true;while (!queue.isempty ()) {temjudge=queue.poll (); List.add ( Temjudge); temj=temjudge%col;//column Position TeMi= (TEMJUDGE-TEMJ)/col;//line position if (temi==0| | temj==0| | temi==row-1| | TEMJ==COL-1) label=false;//If the node is at the boundary, this is not surrounded, so all the nodes that are traversed do not change, here is a label to record if (temj>0&&board[temi][temj-1 ]== ' O ' &&!visited[temi][temj-1])//left {Queue.add (temi*col+temj-1); visited[temi][temj-1]=true;} if (temi>0&&board[temi-1][temj]== ' O ' &&!visited[temi-1][temj])//on {Queue.add (temi-1) *COL+TEMJ ); visited[temi-1][temj]=true;} if (temj<board[0].length-1&&board[temi][temj+1]== ' O ' &&!visited[temi][temj+1])//Right {Queue.add ( temi*col+temj+1); visited[temi][temj+1]=true;} if (temi<board.length-1&&board[temi+1][temj]== ' O ' &&!visited[temi+1][temj])//under {Queue.add ( temi+1) *col+temj); visited[temi+1][temj]=true;}} If the (label)//traversal of all the nodes is not at the boundary, this is a change to these nodes {for (int k=0;k<list.size (); k++) {temjudge=list.get (k); temj=temjudge%    Col    Temi= (TEMJUDGE-TEMJ)/col;board[temi][temj]= ' X ';}} }}</span>`

[Leetcode] [Java] Surrounded regions

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