Surrounded Regions

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

A region is captured by flipping all 'O's into 'X's in that surrounded region.

For example,

X X X XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

题目解析：

反向思索最简单：哪些‘O’是应该保留的？ 从上下左右四个边界往里走，凡是能碰到的‘O’ ，都是跟边界接壤的，应该保留。 思路： 对于每一个边界上的‘O’作为起点，做若干次广度优先搜索，对于碰到的‘O’，标记为其他某字符Y； 最后遍历一遍整个地图，把所有的Y恢复成‘O’，把所有现有的‘O’都改成‘X’。

public class Solution {    static class Pair{        public int first;        public int second;        public Pair(int f, int s){            first = f;            second = s;        }    }     public static boolean  isValid(Pair p, char[][] board){        int m= board.length;        int n= board[0].length;        int x = p.first;        int y = p.second;        if(x<0 || x>=m || y<0 ||y >=n || board[x][y]!='O')             return false;        return true;    }        public static List<Pair> state_extend(Pair pr, char[][] board){        int m= board.length;        int n= board[0].length;        List<Pair> result = new ArrayList<Pair>();        int x = pr.first;        int y = pr.second;                //上下左右        Pair[] new_states = {new Pair(x-1,y),new Pair(x+1,y),                              new Pair(x,y+1),new Pair(x,y-1)};                for(int k =0; k<4; ++k){            if(isValid(new_states[k],board)){                board[new_states[k].first][new_states[k].second] ='+';                result.add(new_states[k]);            }        }                return result;    }                public void solve(char[][] board) {        if(board.length ==0 )return ;                //m,n赋值         int m = board.length;         int n = board[0].length;                for(int i = 0; i<n; i++){            bfs(board,0,i);            bfs(board,m-1, i);        }                for(int j=1; j<m-1;j++){            bfs(board,j,0);            bfs(board,j,n-1);        }                for(int i=0; i<m;i++){            for(int j =0; j<n; j++){                if( board[i][j] =='O')board[i][j]='X';                else if(board[i][j]=='+')                    board[i][j]='O';            }        }//for            }//solve            public void  bfs(char[][] board, int i, int j){                       ArrayDeque<Pair> stack = new ArrayDeque<>();                int m = board.length;        int n = board[0].length;                Pair p = new Pair(i,j);                if(isValid(p,board)){            board[i][j]='+';            stack.push(p);        }               List<Pair> list = new ArrayList<Pair>();        while(!stack.isEmpty()){            Pair cur = stack.pop();            list = state_extend(cur,board);                        for(Pair s: list) {                stack.push(s);            }         }    }        }

网上方法，相对简洁

import java.util.LinkedList;import java.util.List;/** * Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. *  * A region is captured by flipping all 'O's into 'X's in that surrounded region. * * answer: 从上下左右四个边界找O，然后进行BFS，找到的都是要保留的，最后遍历一次修改字符即可。 * 特别注意有一个用例 board 是空，所以进行判断否则会导致RE *  */public class SurroundedRegions {class Point{int x,y;Point(int i,int j){x=i;y=j;}}    public void solve(char[][] board) {    int m = board.length;    if(m == 0){//判空防止 run-time error，不能和 n 一起判断    return;    }    int n = board[0].length;    if(n == 0){//判空防止 run-time error    return;    }    for(int i=0;i<m;i++){    bfs(board,i,0);    bfs(board,i,n-1);    }    for(int j=0;j<n;j++){    bfs(board,0,j);    bfs(board,m-1,j);    }        for(int i=0;i<m;i++){    for(int j=0;j<n;j++){    if(board[i][j] == 'O'){    board[i][j] = 'X';    }else if(board[i][j] == 'A'){    board[i][j] = 'O';    }    }    }    }        private void bfs(char[][] board,int x,int y){    if(board[x][y] != 'O'){    return ;    }    LinkedList<Point> queue = new LinkedList<>();    queue.add(new Point(x,y));    board[x][y] = 'A';    while(!queue.isEmpty()){    Point top = queue.poll();    int i = top.x;    int j = top.y;    addPoint(board,i-1,j,queue);    addPoint(board,i+1,j,queue);    addPoint(board,i,j-1,queue);    addPoint(board,i,j+1,queue);    }    }        private boolean addPoint(char[][] board,int x,int y,LinkedList<Point> queue){    if(x < 0 || x > board.length-1 || y < 0 || y > board[0].length-1 || board[x][y] != 'O'){    return false;    }else{        queue.add(new Point(x,y));        board[x][y] = 'A';    return true;    }    }}