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.."]
]

这道题就是求解n皇后问题。皇后不能出在同一条直线、竖线以及斜线上。

想法就是　　1、建立一个n的数组pos[n]，表示第i行的皇后放在第pos[i]处。

　　　　　　2、然后需要在第num个放皇后的时候，需要依次判断前num个数字（i :  0 --> num-1），这个位置的数字不能与pos[i]，pos[i]-(num-i)，pos[i]+(num-i)这三个数字相同，判断完之前的所有数字之后，就可以放入该数字了。

结果是3ms。算是还可以的成绩。

public class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> result = new ArrayList<List<String>>();
        int[] pos = new int[n];
        putQueen(pos,0,result);
        return result;
    }
	public void putQueen(int[] pos,int num,List<List<String>> result){
		if( num == pos.length ){
			List<String> res = new ArrayList<String>();
			char[] ans = new char[pos.length];
			for( int j =0;j<pos.length ;j++)
					ans[j] = '.';
			for( int i = 0;i<pos.length; i++ ){
				ans[pos[i]] = 'Q';
				res.add(new String(ans));
				ans[pos[i]] = '.';
			}
			result.add(res);
		}
		int[] exi = new int[pos.length];
		for( int i = 0;i<num;i++){
			exi[pos[i]] = 1;
			if( pos[i]-(num-i) >= 0)
				exi[pos[i]-(num-i)] = 1;
			if( pos[i]+num-i < pos.length)
				exi[pos[i]+num-i] = 1;
		}
		
		for( int i = 0; i<pos.length;i++){
			if( exi[i] == 0){
			    pos[num] = i;
				putQueen(pos,num+1,result);
			}
		}	
	}
}

52题

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

这题和上面的一样，就是求出解的个数。

代码就是简单修改一下即可。

public class Solution {
    public int totalNQueens(int n) {
        int[] pos = new int[n];
        return putQueen(pos,0,0);
    }
	public int putQueen(int[] pos,int num,int total){
		if( num == pos.length ){
			return total+1;
		}
		int[] exi = new int[pos.length];
		for( int i = 0;i<num;i++){
			exi[pos[i]] = 1;
			if( pos[i]-(num-i) >= 0)
				exi[pos[i]-(num-i)] = 1;
			if( pos[i]+num-i < pos.length)
				exi[pos[i]+num-i] = 1;
		}
		
		for( int i = 0; i<pos.length;i++){
			if( exi[i] == 0){
			    pos[num] = i;
				total = putQueen(pos,num+1,total);
			}
		}	
		return total;
	}
}