N-Queens

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.."]
]

Java代码：

public class Solution {
    private  int[] column;
	private  int[] rup;
	private  int[] lup;
	private  int[] queen;
	private  List<String[]> result_list;

	public  List<String[]> solveNQueens(int n) {
		Init(n);
		backtrack(1, n);
		return result_list;
	}

	public  void Init(int n) {
		column = new int[n + 1];
		rup = new int[(2 * n) + 1];
		lup = new int[(2 * n) + 1];

		for (int i = 1; i <= n; i++)
			column[i] = 1;

		for (int i = 1; i <= (2 * n); i++)
			rup[i] = lup[i] = 1;

		queen = new int[n + 1];
		result_list = new ArrayList<String[]>();
	}

	public  void backtrack(int i, int num) {
		if (i > num) {
			String[] result = getStringArray(num);
			result_list.add(result);
		} else {
			for (int j = 1; j <= num; j++) {
				if ((column[j] == 1) && (rup[i + j] == 1)
						&& (lup[i - j + num] == 1)) {
					queen[i] = j;
					// 设定为占用
					column[j] = rup[i + j] = lup[i - j + num] = 0;
					backtrack(i + 1, num);
					column[j] = rup[i + j] = lup[i - j + num] = 1;
				}
			}
		}
	}

	public  String[] getStringArray(int num) {
		String[] result = new String[num];
		StringBuilder sb = new StringBuilder();
		for (int y = 1; y <= num; y++) {
			for (int x = 1; x <= num; x++) {
				if (queen[y] == x) {
					sb.append("Q");
				} else {
					sb.append(".");
				}
			}
			result[y - 1] = sb.toString();
			sb.setLength(0);
		}
		return result;
	}
}