皇后问题的经典做法

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<Integer> occupiedCols = new HashSet<Integer>(); private final Set<Integer> occupiedDiag1s = new HashSet<Integer>(); private final Set<Integer> occupiedDiag2s = new HashSet<Integer>(); 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 hereif (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; }

 

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