<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 {
public:
    vector<vector<string>> ans;
    string temp;

    void dfs(vector<string> now, vector<vector<bool>> vis, int z, int n){
        if(z==n){ans.push_back(now);return;}
        for(int i=0;i<n;i++){
            if(vis[z][i])continue;
            string s = temp;
            s[i]='Q';
            vector<vector<bool>> vis2=vis;
            for(int x=0;x<n;x++){
                vis[z][x]=true;
                vis[x][i]=true;
                if(i+x<n && z+x<n)vis[z+x][i+x]=true;
                if(z+x<n && i-x>=0)vis[z+x][i-x]=true;
            }
            now.push_back(s);
            dfs(now, vis, z+1, n);
            now.pop_back();
            vis=vis2;
        }
    }

    vector<vector<string>> solveNQueens(int n) {
        for(int i=0;i<n;i++)temp.append(".");
        for(int i=0;i<n;i++){
            vector<vector<bool>> vis(n, vector<bool>(n, false));
            string s=temp;
            s[i]='Q';
            for(int x=0;x<n;x++){
                vis[0][x]=true;
                vis[x][i]=true;
                if(i+x<n)vis[x][i+x]=true;
                if(i-x>=0)vis[x][i-x]=true;
            }
            dfs({s}, vis, 1, n);
        }
        return ans;
    }
};
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