# N Queens

## Question

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.

Have you met this question in a real interview? Yes

**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?

## Analysis

// TODO Add more 渐进式方法

## Solution

```java
import java.util.Arrays;
import java.util.ArrayList;

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) {
        ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
        String[] nQueens = new String[n];
        char[] init = new char[n];
        Arrays.fill(init, '.');
        Arrays.fill(nQueens, String.valueOf(Arrays.copyOf(init, n)));
        search(n, 0, nQueens, result);
        return result;
    }


    private void search(int n, int row, String[] nQueens, ArrayList<ArrayList<String>> result) {
        if (row == n) {
            result.add(new ArrayList<String>(Arrays.asList(nQueens)));
            return;
        }

        for (int col = 0; col < n; col++) {
            if (isValid(nQueens, row, col, n)) {
                char[] chars;
                chars = nQueens[row].toCharArray();
                chars[col] = 'Q';
                nQueens[row] = String.valueOf(chars);
                // nQueens[row][col] = 'Q';
                search(n, row + 1, nQueens, result);
                chars = nQueens[row].toCharArray();
                chars[col] = '.';
                nQueens[row] = String.valueOf(chars);
                // nQueens[row][col] = '.';
            }
        }
    }

    private boolean isValid(String[] nQueens, int row, int col, int n) {
        char[] chars;
        for (int i = 0; i < row; i++) {
            chars = nQueens[i].toCharArray();
            if (chars[col] == 'Q') {
                return false;
            }
        }

        for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {
            chars = nQueens[i].toCharArray();
            if (chars[j] == 'Q') {
                return false;
            }
        }

        for (int  i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) {
            chars = nQueens[i].toCharArray();
            if (chars[j] == 'Q') {
                return false;
            }
        }
        return true;
    }

    public static void main(String[] args) {
        Solution s = new Solution();
        ArrayList<ArrayList<String>> res = s.solveNQueens(Integer.valueOf(args[0]));
        // java Solution 4
        System.out.println(res);
        //[[.Q.., ...Q, Q..., ..Q.], [..Q., Q..., ...Q, .Q..]]
    }
};
```

## Reference

* [wikipedia: Eight Queens Puzzle](https://en.wikipedia.org/wiki/Eight_queens_puzzle)
* [LeetCode “全排列”问题系列(一) - 用交换元素法生成全排列及其应用，例题: Permutations I 和 II, N-Queens I 和 II，数独问题](http://www.cnblogs.com/felixfang/p/3705754.html)


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