> For the complete documentation index, see [llms.txt](https://kongchengzhuge.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://kongchengzhuge.gitbook.io/leetcode/51.-n-queens.md).

# 51. N-Queens

## problem description

The *n*-queens puzzle is the problem of placing *n* queens on an *n*×*n* chessboard such that no two queens attack each other.

![](https://assets.leetcode.com/uploads/2018/10/12/8-queens.png)

Given an integer *n*, return all distinct solutions to the *n*-queens puzzle.

Each solution contains a distinct board configuration of the *n*-queens' placement, where `'Q'` and `'.'` both indicate a queen and an empty space respectively.

**Example:**

```
Input: 4
Output: [
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
```

## algorithm thought

经典的n皇后问题。使用回溯法解决问题，每次放之前首先判断这个位置是否合法，如果合法，放入这位置。如果不合法判断下个能放的位置。最后如果没有能放的位置的时候，就回溯。

## code

```cpp
class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        vector<vector<string>> res;
        string t(n,'.');
        vector<string> tmp;
        for(int i=0;i<n;++i)
            tmp.push_back(t);
        helper(res,tmp,0);
        return res;
    }
    void helper(vector<vector<string>>& res,vector<string> tmp,int pos){
        if(pos==tmp.size()){
            res.push_back(tmp);
            return;
        }
        for(int i=0;i<tmp.size();++i){
            if(isvalid(pos,i,tmp)){
                tmp[pos][i]='Q';  
                helper(res,tmp,pos+1);
                tmp[pos][i]='.';    
            }  
        }
    }
    bool isvalid(int x,int y,vector<string> res){
        for(int i=0;i<res.size();++i){
            if(res[x][i]=='Q'||res[i][y]=='Q')
                return false;
        }
        int i=x-1,j=y-1;
        while(i>=0&&j>=0){
            if(res[i--][j--]=='Q')
                return false;
        }
        i=x-1;j=y+1;
        while(i>=0&&j<res.size()){
            if(res[i--][j++]=='Q')
                return false;
        }
        return true;
    }
};
```

## algorithm analysis

这里的时间复杂度我也不是太清楚，查了资料，网上有两种说法，我觉得O(n^n)和O(n!)都有道理。