lintcode-medium-N Queens

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

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
;

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;
}

};