LeetCode in Python 51. N-Queens

Posted 2022-04-18 lowkeysingsing

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

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.

Solution:

(Python)

class Solution(object):
    def solveNQueens(self, n):
        """
        :type n: int
        :rtype: List[List[str]]
        """
        res = []
        def dfs(rcSum, rcDiff, queens):
            if len(queens) == n:
                queens = [‘.‘ * i + ‘Q‘ + ‘.‘ * (n-i-1) for i in queens]
                res.append(queens)
                return
            row = len(queens)
            for col in range(n):
                if col in queens or row+col in rcSum or row-col in rcDiff: continue
                dfs(rcSum+[row+col], rcDiff+[row-col], queens+[col])
        
        dfs([], [], [])
        return res

(C++)

class Solution 
public:
    vector<vector<string>> solveNQueens(int n) 
        vector<vector<string>> res;
        vector<string> out (n, string(n, ‘.‘));
        dfs(n, res, out, 0);
        return res;
    
    void dfs(int n, vector<vector<string>>& res, vector<string>& out, int row) 
        if (row == n) 
            res.push_back(out);
            return;
        
        for (int col = 0; col < n; col++) 
            out[row][col] = ‘Q‘;
            if (isValid(n, row, col, out)) 
                dfs(n, res, out, row+1);
            out[row][col] = ‘.‘;
        
    
    bool isValid(int n, int row, int col, vector<string>& out) 
        // 横向改变Queen，因此不需要检查水平方向
        
        // vertical
        for (int i = 0; i < row; i++)
            if (out[i][col] == ‘Q‘) return false;
        
        // left diagonal
        for (int i = row-1, j = col-1; i >= 0 && j >= 0; i--, j--)
            if (out[i][j] == ‘Q‘) return false;
        
        // right diagonal
        for (int i = row-1, j = col+1; i >= 0 && j < n; i--, j++)
            if (out[i][j] == ‘Q‘) return false;
        
        return true;
    
;

Python版相比C++更简洁，用queens[i]记录第i+1行的皇后所在的列序号。经典for循环中调dfs递归。注意Python解法用了这样一种原理：row+col或row-col相等，对角线可达。