i for i in listrow if i != ''

This is a list comprehension in Python that loops through each element in the list "listrow" and checks if it is not an empty string (""). If the element is not an empty string, it is included in a new list that is created by the list comprehension.

For example, if "listrow" was ["hello", "", "world"], the list comprehension would return ["hello", "world"].

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listrow = [i for i in listrow if i != '']

This line of code creates a new list called "listrow" by iterating through each element in an existing list (not shown in the code) and adding it to the new list only if it's not an empty string. The resulting list will contain all the non-empty elements from the original list.

For example, if the original list was ["dog", "", "cat", "", "bird"], the code would create a new list with the values ["dog", "cat", "bird"].

class Node: def __init__(self, board=None, row=None, col=None): self.board = board self.row = row self.col = col self.moves = [] def get_moves(self): last_row = self.row for col in range(len(self.board)): if is_safe(self.board, last_row + 1, col): new_board = [row[:] for row in self.board] new_board[last_row + 1][col] = 1 self.moves.append(Node(new_board, last_row + 1, col)) def is_safe(board, row, col): for i in range(row): if board[i][col] == 1: return False for i, j in zip(range(row, -1, -1), range(col, -1, -1)): if board[i][j] == 1: return False for i, j in zip(range(row, -1, -1), range(col, len(board), -1)): if board[i][j] == 1: return False return True def solve_n_queens(n): root = Node([[0 for _ in range(n)] for _ in range(n)], -1, None) current_list = [root] while current_list: node = current_list.pop(0) if node.row == n - 1: return node.board node.get_moves() for move in node.moves: current_list.append(move) n = int(input("请输入皇后数量：")) solution = solve_n_queens(n) for row in solution: print(row)该代码用的算法是什么

这段代码使用了经典的 N 皇后问题的解法，采用了深度优先搜索算法。具体地，代码中定义了一个 `Node` 类来表示棋盘上的一种情况，每个 `Node` 实例都代表棋盘上的一行。在 `get_moves` 方法中，对于当前行的每一个可行位置，都创建一个新的 `Node` 实例来代表下一行的情况，并将其添加到 `moves` 列表中。然后在主函数 `solve_n_queens` 中，使用一个队列来记录当前的搜索路径，不断从队列中取出 `Node` 实例，直到找到一个满足条件的解或者队列为空。具体来说，在每次取出一个 `Node` 实例时，如果它所代表的行是最后一行，则说明找到了一个解，直接返回该实例所代表的棋盘状态；否则，调用 `get_moves` 方法得到下一行的所有可行位置，并将对应的 `Node` 实例添加到队列中。由于搜索过程中，每个 `Node` 实例都只会被访问一次，因此该算法的时间复杂度为 $O(n^n)$，空间复杂度为 $O(n^2)$。