N-Queens

Q:
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.."]
]

NOTE: The queen (♕,♛) is the most powerful piece in the game of chess, 

able to move any number of squares vertically, horizontally, or diagonally. 

A:
 --------------------2 nd pass----------------------
DFS ----------  Use 2D array
出现的问题，还是在于，测试diagonal的时候， 两条diagonal的表达有问题。比较坐标系不同。而这个问题其实很好解决。就是看(row,col) 所在的对角线有什么特征即可。不用考虑坐标系转换。

public class Solution { 
    public List<List<String>> solveNQueens(int n) {
        char[][] board = new char[n][n];
        for(int i=0;i<n;i++)
            Arrays.fill(board[i],'.');
        List<List<String>> all = new LinkedList<List<String>>();
        dfs(board,0,all);
        return all;
    }
    private void dfs(char[][] board, int col, List<List<String>> all){
        for(int i =0;i<board.length;i++){
            board[i][col] = 'Q';
            if(isValid(board,i,col)){
                if(col==board.length-1){// done
                    List list = new LinkedList<String>();
                    for(int j =0;j<board.length;j++)
                        list.add(new String(board[j]));
                    all.add(list);
                }else{
                    dfs(board,col+1,all);
                }
            }
            board[i][col] = '.';
        }
    }
    private boolean isValid(char[][] board, int row ,int col){ // board[i][j] is just been set
        for(int j=0; j<col; j++)
            if(board[row][j]=='Q')
                return false;
        for(int i=0;i<board.length;i++){ 
            int j = row + col -i;// check diagal /
            if(i!= row && j>=0 && j< col && board[i][j]=='Q')
                return false;
            j = i + col - row; // check diagonal \
            if(i!= row && j>=0 && j< col && board[i][j]=='Q')
                return false;
        }
        return true;
    }
}

--------------3rd pass---------------------

这里用了2D array，  如果用一维数组表示的话（类似于N-Queen II) ，

方法是：一维数组的下标表示横坐标（哪一行），而数组的值表示纵坐标（哪一列）。

Mistakes:
1：  System.out.print(str[i].charAt(j) + '\t'); 打印出的结果是数字。  哎，  不是char 类型
应该写成System.out.print(""+str[i].charAt(j) + '\t');     加个"" 就好了。
2：