N-Queens

题目：

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.

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

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

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]
  
思路： 这题就像上篇我们做的palindrome partition是一个类型的。等着把这类题写在一起做个对比。先直接上代码：

    public List<String[]> solveNQueens(int n) {
        List<String[]> ret = new ArrayList<String[]>();
        if (n == 0) {
            return ret;
        }
        helper(n, 0, new int[n], ret);
        return ret;
    }
    
    private static void helper(int n, int row, int[] marked, List<String[]> ret) {
        if (row == n) {
            String[] subRet = new String[n];
            for (int i = 0; i < n; i++) {
                StringBuilder str = new StringBuilder();
                for (int j = 0; j < n; j++) {
                    if (marked[i] == j) {
                        str.append('Q');
                    }else{
                        str.append('.');
                    }
                }
                subRet[i] = str.toString();
            }
            ret.add(subRet);
            //forget to return;
            return;
        }
        for (int i = 0; i < n; i++) {
            marked[row] = i
            if (isValid(row, marked)) {
                helper(n, row + 1, marked, ret);
            }
        }
    }
    
    private static boolean isValid(int row, int[] marked) {
        for (int i = 0; i < row; i++) {
            if (marked[row] == marked[i] || Math.abs(marked[row] - marked[i]) = )
        }
    }