leetcode: 51. N-Queens

# The n-queens puzzle is the problem of placing n queens on

# an nxn chess board such that no two queens attack each other.
# 
# Given an integer n, return all distinct solutions to the n-queens puzzle.
# 
# Each solution contains a distinct board configuration of the n-queens' placement, 
# where 'Q' and '.' both indicate a queen and an empty space respectively.
# 
# For example,
# There exist two distinct solutions to the 4-queens puzzle:
# 
# [
#  [".Q..",  // Solution 1
#   "...Q",
#   "Q...",
#   "..Q."],
# 
#  ["..Q.",  // Solution 2
#   "Q...",
#   "...Q",
#   ".Q.."]
# ]

1. 逐行放置皇后：排除在同一行的可能。
2. 记录之前所放皇后的列坐标：col[i]=j表示第i行的皇后在第j列。这样在放置第i+1行时，只要保证col[i+1] != col[k], k=0...i 即可。
3. 对角线判断：对于任意(i1, col[i1])和(i2, col[i2])，只有当abs(i1-i2) = abs(col[i1]-col[i2])时，两皇后才在同一对角线。

class Solution():
    def solveNQueens(self, n):
        from functools import reduce
        def solveNQueensRecu(solution, row, n):
            if row == n:
                solutions.append(list(map(lambda x: '.' * x + "Q" + '.' * (n - x - 1), solution)))
            else:
                for i in range(n):
                    if i not in solution and reduce(lambda acc, j: abs(row - j) != abs(i - solution[j]) and acc, range(len(solution)), True):
                        solveNQueensRecu(solution + [i], row + 1, n)
        solutions = []
        solveNQueensRecu([], 0, n)
        return solutions


if __name__ == "__main__":
    assert Solution().solveNQueens(4) == [['.Q..', '...Q', 'Q...', '..Q.'], ['..Q.', 'Q...', '...Q', '.Q..']]