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LeetCode——surrounded-regions

作者:互联网

Q:现在有一个仅包含‘X’和‘O’的二维板,请捕获所有的被‘X’包围的区域
捕获一个被包围区域的方法是将被包围区域中的所有‘O’变成‘X’
例如
X X X X↵X O O X↵X X O X↵X O X X
执行完你给出的函数以后,这个二维板应该变成:
X X X X↵X X X X↵X X X X↵X O X X

A:
1.与边界上面的O相连的那些O不是被包围的,所以利用深度搜索将这些与边界相连(包括边界)的O变成
2.剩下的O就是被包围的,将他们改成X
3.再将
变回O

public int rowNum = 0;
public int colNum = 0;
public void solve(char[][] board) {
    if(board == null || board.length <= 0|| board[0].length <= 0){
        return;
    }
    rowNum = board.length;
    colNum = board[0].length;
    for(int i = 0; i < colNum; i++){
        dfs(board, 0, i);
        dfs(board, rowNum-1, i);
    }
    for(int i = 0; i < rowNum; i++){
        dfs(board, i, 0);
        dfs(board, i, colNum-1);
    }
    for(int i = 0; i < rowNum; i++){
        for(int j = 0; j < colNum; j++){
            if(board[i][j] == 'O'){
                board[i][j] = 'X';
            }
        }
    }
    for(int i = 0; i < rowNum; i++){
        for(int j = 0; j < colNum; j++){
            if(board[i][j] == '*'){
                board[i][j] = 'O';
            }
        }
    }
}
private void dfs(char[][] board, int row, int col) {
    if(board[row][col] == 'O'){
        board[row][col] = '*';
        if(row > 1){
            dfs(board, row-1, col);
        }
        if(col > 1){
            dfs(board, row, col-1);
        }
        if(row < rowNum-1){
            dfs(board, row+1, col);
        }
        if(col < colNum-1){
            dfs(board, row, col+1);
        }
    }
}

标签:surrounded,dfs,LeetCode,regions,public,board,包围,col,row
来源: https://www.cnblogs.com/xym4869/p/12488344.html