【leetcode】被围绕的区域

/*深度优先*/
void dfs(char** board,int r,int c,int row,int col)
{
    if (r<0 || r>row-1 || c<0 || c>col-1 || board[r][c] != 'O') return;    
    board[r][c] = 'A';
    dfs(board,r+1,c,row,col);
    dfs(board,r-1,c,row,col);
    dfs(board,r,c+1,row,col);
    dfs(board,r,c-1,row,col);
}

void solve(char** board, int boardSize, int* boardColSize){
    int row=boardSize,col=*boardColSize,i,j;
    if(!row) return;
    for (i=0; i<row; i++)
    {
        dfs(board,i,0,row,col);
        dfs(board,i,col-1,row,col);
    }
    for (i=1; i<col-1; i++)
    {
        dfs(board,0,i,row,col);
        dfs(board,row-1,i,row,col);
    }
    for (i=0; i<row; i++)
    {
        for (j=0; j<col; j++)
        {
            if (board[i][j] == 'A') board[i][j] = 'O';
            else if (board[i][j] == 'O') board[i][j] = 'X';
        }
    }
}
/*广度优先*/
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};

void solve(char** board, int boardSize, int* boardColSize) {
    int n = boardSize;
    if (n == 0) {
        return;
    }
    int m = boardColSize[0];

    int** que = (int**)malloc(sizeof(int*) * n * m);
    for (int i = 0; i < n * m; i++) {
        que[i] = (int*)malloc(sizeof(int) * 2);
    }
    int l = 0, r = 0;
    for (int i = 0; i < n; i++) {
        if (board[i][0] == 'O') {
            board[i][0] = 'A';
            que[r][0] = i, que[r++][1] = 0;
        }
        if (board[i][m - 1] == 'O') {
            board[i][m - 1] = 'A';
            que[r][0] = i, que[r++][1] = m - 1;
        }
    }
    for (int i = 1; i < m - 1; i++) {
        if (board[0][i] == 'O') {
            board[0][i] = 'A';
            que[r][0] = 0, que[r++][1] = i;
        }
        if (board[n - 1][i] == 'O') {
            board[n - 1][i] = 'A';
            que[r][0] = n - 1, que[r++][1] = i;
        }
    }
    while (l < r) {
        int x = que[l][0], y = que[l][1];
        l++;
        for (int i = 0; i < 4; i++) {
            int mx = x + dx[i], my = y + dy[i];
            if (mx < 0 || my < 0 || mx >= n || my >= m || board[mx][my] != 'O') {
                continue;
            }
            board[mx][my] = 'A';
            que[r][0] = mx, que[r++][1] = my;
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (board[i][j] == 'A') {
                board[i][j] = 'O';
            } else if (board[i][j] == 'O') {
                board[i][j] = 'X';
            }
        }
    }
    for (int i = 0; i < n * m; i++) {
        free(que[i]);
    }
    free(que);
}
原文地址:https://www.cnblogs.com/ganxiang/p/13684293.html