leetcode——130.被围绕的区域

做出来了，但是时间用的比较长。

用了递归。

public void solve(char[][] board) {
        int m = board.length;
        if(m == 0){
            return;
        }
        int n = board[0].length;
        if(n == 0){
            return;
        }
        boolean[][] visited = new boolean[m][n];  //标记是否被访问过
        List<List<Integer>> list = new ArrayList<>();  //存放相邻的'O'字符的位置
        for(int i = 0;i<m;i++){
            for(int j = 0;j<n;j++){
                if(!visited[i][j]) {
                    if (board[i][j] == 'X') {
                        visited[i][j] = true;
                    } else {
                        boolean b = true;
                        if(find(i,j,m,n,board,visited,list,b)){     //被包围
                            for (List<Integer> l : list) {
                                board[l.get(0)][l.get(1)] = 'X';
                            }
                        }
                        list.clear();
                    }
                }
            }
        }
    }

    private boolean find(int i, int j, int m, int n, char[][] board, boolean[][] visited, List<List<Integer>> list,boolean b) {
        if(i<0 || i>=m || j<0 || j>= n){
            return b;
        }
        if(!visited[i][j] ) {
            if (board[i][j] == 'O') {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {   //位置(i,j)处的字符是O     //如何判断边界
                    b = false;
                }
                List<Integer> local = new ArrayList<>();
                local.add(i);
                local.add(j);
                list.add(local);    //先将起点坐标存入
                visited[i][j] = true;
                return find(i, j - 1, m, n, board, visited, list, b)
                        & find(i, j + 1, m, n, board, visited, list, b)
                        & find(i + 1, j, m, n, board, visited, list, b)
                        & find(i - 1, j, m, n, board, visited, list, b);
            }else{
                visited[i][j] = true;
            }
        }
        return b;
    }

别人的题解，比我的优化多了

public void solve(char[][] board) {
        if(board.length==0) return;
        int m =board.length,n =board[0].length;
        boolean[][] visit=new boolean[m][n];
        for (int i = 0; i < n; i++) {
            if(board[0][i]=='O'){
                solveHelp(board,0,i,visit);
            }
            if(board[m-1][i]=='O'){
                solveHelp(board,m-1,i,visit);
            }
        }
        for (int i = 0; i < m; i++) {
            if(board[i][0]=='O'){
                solveHelp(board,i,0,visit);
            }
            if(board[i][n-1]=='O'){
                solveHelp(board,i,n-1,visit);
            }
        }
        for (int i = 0; i <m ; i++) {
            for (int j = 0; j <n ; j++) {
                if(board[i][j]=='T') board[i][j]='O';
                else board[i][j]='X';
            }
        }

    }
    void solveHelp(char[][] board,int i,int j,boolean[][] visit){
        if(visit[i][j]) return;
        visit[i][j]=true;
        if(board[i][j]=='O') board[i][j]='T';
        if(i>0&&board[i-1][j]=='O'){
            solveHelp(board,i-1,j,visit);
        }
        if(i<board.length-1&&board[i+1][j]=='O'){
            solveHelp(board,i+1,j,visit);
        }
        if(j>0&&board[i][j-1]=='O'){
            solveHelp(board,i,j-1,visit);
        }
        if(j<board[0].length-1&&board[i][j+1]=='O'){
            solveHelp(board,i,j+1,visit);
        }
    }

 ——2020.8.3