[Leetcode] 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.

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

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

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

Solution:
大神的做法：
http://blog.csdn.net/xift810/article/details/22187633

 1 public class Solution {
 2     public List<String[]> solveNQueens(int n) {
 3     List<String[]> result = new ArrayList<String[]>();
 4         if (n == 0)
 5             return result;
 6         int[] rCol = new int[n];
 7         queens(result, rCol, 0, n);
 8         return result;
 9     }
10 
11     private void queens(List<String[]> result, int[] rCol, int row, int n) {
12         // TODO Auto-generated method stub
13         if (row == n) {
14             printQueens(result, rCol, n);
15             return;
16         }
17         for (int col = 0; col < n; ++col) {
18             rCol[row] = col;
19             if (check(rCol, row)) {
20                 queens(result, rCol, row + 1, n);
21             }
22         }
23     }
24 
25     private boolean check(int[] rCol, int row) {
26         // TODO Auto-generated method stub
27         for (int i = 0; i < row; ++i) {
28             if (rCol[i] == rCol[row])
29                 return false;
30             if (Math.abs(row - i) == Math.abs(rCol[row] - rCol[i])) {
31                 return false;
32             }
33         }
34         return true;
35     }
36 
37     private void printQueens(List<String[]> result, int[] rCol, int n) {
38         // TODO Auto-generated method stub
39         String[] s = new String[n];
40         for (int i = 0; i < n; ++i) {
41             String temp = "";
42             for (int j = 0; j < n; ++j) {
43                 if(rCol[i]==j){
44                     temp+="Q";
45                 }else{
46                     temp+=".";
47                 }
48             }
49             s[i]=temp;
50         }
51         result.add(s);
52     }
53 }