*LeetCode-N-Queens

backtrack的题就是不敢确定是否可以这样做 这个题的细节就是如何check是否这个格子可以放q 需要一个check function

还有就是先循环行 还是列 

其实外层循环是在循环列 在同一行中试图放在不同列中 然后放了一个之后进入下一层recursion 放下一行

同时要注意使用string builder

public class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> res = new ArrayList<List<String>> ();
        List<String> list = new ArrayList<String> ();
        helper ( res, list, 0, n );
        return res;
    }
    public void helper ( List<List<String>> res, List<String> list, int row, int n ){
        if ( row == n ){
            res.add( new ArrayList<String>(list));
            return;
        }
        for ( int i = 0; i < n; i ++ ){
            StringBuilder thisRow = new StringBuilder();
            for ( int j = 0; j < n; j ++ ){
                thisRow.append('.');
            }
            if ( valid ( list, i, n )){
                thisRow.setCharAt( i, 'Q');
                list.add( thisRow.toString() );
                helper ( res, list, row + 1, n);
                list.remove( list.size() - 1 );
            }
        }
    }
    
    public boolean valid ( List<String> list, int col, int n ){
        for ( int i = 0; i < list.size(); i ++ ){
            if ( list.get( i ).charAt(col) == 'Q' )
                return false;
        }
        for ( int i = list.size() - 1; i >= 0; i -- ){
            if ( (list.size() - i + col) < n && list.get(i).charAt(list.size() - i + col) == 'Q')
                return false;
        }
        for ( int i = list.size() - 1; i >= 0; i -- ){
            if ( (col - list.size() + i) >= 0 && list.get(i).charAt(col - list.size() + i) == 'Q')
                return false;
        }
        return true;
    }
}