Question: https://oj.leetcode.com/problems/n-queens-ii/

Question Name: N-Queens II

Same method as that in N-Queens problem. But we could save some time and space by not recording the chessboard.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 | class QueenSolution(object): ''' A class to store the intermediate status towards the N-Queen solutions ''' __slots__ = ("col","diagF", "diagB") def __init__(self, col, diagF, diagB): self.col = col # Used columns self.diagF = diagF # Used forward diagonal self.diagB = diagB # Used backward diagonal return class Solution: # @return an integer def totalNQueens(self, n): # Handle some special cases if n == 1: return 1 if n == 2 or n == 3: return 0 result = [QueenSolution(0, 0, 0)] for row in xrange(n): # Add the queens row by row temp = [] for preState in result: for col in xrange(n): # Try each position # Column conflicts if (1<<col) & preState.col != 0: continue diagF = 1 << (col - row + (n-1)) # Forward diagonal conflicts if diagF & preState.diagF != 0: continue diagB = 1 << (row + col) # Backward diagonal conflicts if diagB & preState.diagB != 0: continue # Find a valid intermediate status temp.append(QueenSolution( (1<<col) | preState.col, diagF | preState.diagF, diagB | preState.diagB)) result = temp # Prepare for working with the next row return len(result) |