<p>The <strong>n-queens</strong> puzzle is the problem of placing <code>n</code> queens on an <code>n x n</code> chessboard such that no two queens attack each other.</p>

<p>Given an integer <code>n</code>, return <em>all distinct solutions to the <strong>n-queens puzzle</strong></em>. You may return the answer in <strong>any order</strong>.</p>

<p>Each solution contains a distinct board configuration of the n-queens&#39; placement, where <code>&#39;Q&#39;</code> and <code>&#39;.&#39;</code> both indicate a queen and an empty space, respectively.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/queens.jpg" style="width: 600px; height: 268px;" />
<pre>
<strong>Input:</strong> n = 4
<strong>Output:</strong> [[&quot;.Q..&quot;,&quot;...Q&quot;,&quot;Q...&quot;,&quot;..Q.&quot;],[&quot;..Q.&quot;,&quot;Q...&quot;,&quot;...Q&quot;,&quot;.Q..&quot;]]
<strong>Explanation:</strong> There exist two distinct solutions to the 4-queens puzzle as shown above
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> [[&quot;Q&quot;]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 9</code></li>
</ul>

class Solution:
    def solveNQueens(self, n: int) -> List[List[str]]:
        #这题刷的少 咋一看没思路 但重新看下一下就想起来了
        self.res = []  
        self.board = [['.']*n for _ in range(n)]
        self.d = set()  #正对角线 row-col [4,4]和[0,0]在同一条正对角线上
        self.anti_d = set() #反对角线 row+col [1,3]和[3,1]在同一条反对角线上
        self.col = set()  #【这个是最费解的】但下面dfs是生成每一个row 这里单独记录col也是可以理解的
        self.dfs(0, n) #row_number, n
        return self.res

    def dfs(self, row, n):
        if row == n: #row不能等于n-1 必须走完最后一行 到它后面
            return self.res.append([''.join(row) for row in self.board]) #【忘记了[]】

        for col in range(n): #col是row里的index
            diff = row - col
            Sum = row + col 
            if not (col in self.col or diff in self.d or Sum in self.anti_d):
                self.col.add(col) #同一列里没有相同数字 
                self.d.add(diff)  
                self.anti_d.add(Sum)  
                self.board[row][col] = 'Q'  
                self.dfs(row+1, n)   

                self.col.remove(col) #同一列里没有相同数字 
                self.d.remove(diff)  
                self.anti_d.remove(Sum)  
                self.board[row][col] = '.'  

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