lintcode-medium-N Queens

The n-queens puzzle is the problem of placing n queens on an n×nchessboard 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

There exist two distinct solutions to the 4-queens puzzle:

[
  // Solution 1
  [".Q..",
   "...Q",
   "Q...",
   "..Q."
  ],
  // Solution 2
  ["..Q.",
   "Q...",
   "...Q",
   ".Q.."
  ]
]

Challenge

Can you do it without recursion?

 

class Solution {
    /**
     * Get all distinct N-Queen solutions
     * @param n: The number of queens
     * @return: All distinct solutions
     * For example, A string '...Q' shows a queen on forth position
     */
    ArrayList<ArrayList<String>> solveNQueens(int n) {
        // write your code here
        
        ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
        
        if(n <= 0) 
            return result;
        
        int[] position = new int[n];
        
        helper(result, 0, position, n);
        
        return result;
    }
    
    public void helper(ArrayList<ArrayList<String>> result, int row, int[] position, int n){
        
        if(row == n){
            ArrayList<String> solution = new ArrayList<String>();
            for(int i = 0; i < n; i++){
                StringBuilder line = new StringBuilder();
                
                for(int j = 0; j < n; j++){
                    if(j == position[i])
                        line.append('Q');
                    else
                        line.append('.');
                }
                
                solution.add(line.toString());
            }
            
            result.add(new ArrayList<String>(solution));
            return;
        }
        
        for(int i = 0; i < n; i++){
            position[row] = i;
            if(valid(position, row, n))
                helper(result, row + 1, position, n);
        }
        
        return;
    }
    
    public boolean valid(int[] position, int row, int n){
        
        for(int i = row - 1; i >= 0; i--){
            if(position[row] == position[i] || Math.abs(position[row] - position[i]) == row - i)
                return false;
        }
        
        return true;
    }
    
};