*LeetCode-N-Queens

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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;    }}

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