求个数，和打印全部，都可以用下面这种非常简洁的代码结构，来做。

 

/** * don't need to actually place the queen, * instead, for each row, try to place without violation on * col/ diagonal1/ diagnol2. * trick: to detect whether 2 positions sit on the same diagnol: * if delta(col, row) equals, same diagnol1; * if sum(col, row) equals, same diagnal2. */private final Set
     
       occupiedCols = new HashSet
      
       ();private final Set
       
         occupiedDiag1s = new HashSet
        
         ();private final Set
         
           occupiedDiag2s = new HashSet
          
           ();public int totalNQueens(int n) { return totalNQueensHelper(0, 0, n);}private int totalNQueensHelper(int row, int count, int n) { for (int col = 0; col < n; col++) { if (occupiedCols.contains(col)) continue; int diag1 = row - col; if (occupiedDiag1s.contains(diag1)) continue; int diag2 = row + col; if (occupiedDiag2s.contains(diag2)) continue; // we can now place a queen here if (row == n-1) count++; else { occupiedCols.add(col); occupiedDiag1s.add(diag1); occupiedDiag2s.add(diag2); count = totalNQueensHelper(row+1, count, n); // recover occupiedCols.remove(col); occupiedDiag1s.remove(diag1); occupiedDiag2s.remove(diag2); } } return count;}