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.."]
]

八皇后问题的变形题，一个不能再经典的题了。

回溯法，要保证随意两各queens都不能再同一行、同一列、同一对角线。代码例如以下：

    public List<List<String>> solveNQueens(int n) {
        List<List<String>> lists = new ArrayList<List<String>>();
        dfs(lists, new int[n], 0);
        return lists;
    }
    
    private void dfs(List<List<String>> lists, int[] ret, int col) {
    	if (col == ret.length) {
    		List<String> list = new ArrayList<String>(ret.length);
    		for (int p = 0; p < ret.length; p++) {
    			StringBuilder sb = new StringBuilder();
    			for (int q = 0; q < ret.length; q++) {
    				if (ret[p] == q) sb.append("Q");
    				else sb.append(".");
    			}
    			list.add(sb.toString());
    		}
    		lists.add(list);
    		return;
    	}
    	for (int k = 0; k < ret.length; k++) {
    		ret[col] = k;
    		if (check(ret, col))
    			dfs(lists, ret, col+1);
    	}
    }
    
    private boolean check(int[] ret, int col) {    	
    	for (int p = 0; p < col; p++) {
    		if (Math.abs(ret[col]-ret[p]) == Math.abs(col-p))
    			return false;
    		if (ret[col] == ret[p])
    			return false;
    	}
    	return true;
    }

事实上这里面还是有非常多值得研究的，首先搞了一个n长度的一维数组，而不是二维数组。由于数组的下标就代表了第几列。即ret[col] = row， 这是一个非常好的优化。我上面是一列一列的扫描的。你也能够一行一行的扫描。都一样。由于这是个n*n的正方形。