LeetCode37题解（yield生成器提高速度）

37.解数独

编写一个程序，通过已填充的空格来解决数独问题。
一个数独的解法需遵循如下规则：

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

空白格用 '.' 表示。（左侧为题目，右侧为一个解答）

Note:

• 给定的数独序列只包含数字 1-9 和字符 '.' 。
• 你可以假设给定的数独只有唯一解。
• 给定数独永远是 9x9 形式的。

解题思路

9×9：看作3×3，然后每个单元又是一个3×3小块
Solution：初始化小块
big_permutations：大块中挑选组合数最少的小块，
    原因在于，一旦小块确定下来，周边的小块的组合数会下降（由于限定条件增加），
    加快搜索速度，确定某个组合数最少的小块，就把数字填入board，
    再从剩余小块筛选组合数最少的小块，以此类推。
possible_blk：每个候选小块（有些小块已经确定，就不在候选范围内）的可能组合，
    取其中组合数最少的返回，为了不浪费，生成n个生成器，不用马上那个取回所有结果，
    而是先定义生成器，然后用next(生成器)来取，谁先取完，就是组合数最少的一个
small_permutations：小块的组合
    比如小块中有4个空白，
    每个空白可选值如下（由于块，行，列已经填入数字的限制）
        [['4'], ['2', '7'], ['2', '7'], ['1']]
    那么可能的组合如下：
        ['4', '2', '7', '1']
        ['4', '7', '2', '1']

代码

class Solution37(object):
    def solveSudoku(self, board):
        blk = [[board[b//3 * 3+ i//3][(b%3)*3+i%3] for i in range(9)]for b in range(9)]
        blank_pos = [[i for i in range(9) if blk[k][i]=='.'] for k in range(9)]
        blank_val = [[str(i) for i in range(1,10) if not str(i) in blk[no]] for no in range(9)]
        for res in self.big_permutations(board,blank_pos,blank_val):
            return res
            
    def big_permutations(self,board,blank_pos,blank_val,b_no=[]):
        no,pp = self.possible_blk(board,blank_pos,blank_val,b_no)
        for p3 in pp:
            for i in range(len(p3)):
                board[no//3*3 + blank_pos[no][i]//3][(no%3)*3 + blank_pos[no][i]%3] = p3[i]
            if len(b_no) == 8:
                yield board[:]
            else:
                for res in self.big_permutations(board,blank_pos,blank_val,b_no + [no]):
                    yield res
            for i in range(len(p3)):
                board[no//3*3 + blank_pos[no][i]//3][(no%3)*3 + blank_pos[no][i]%3] = '.'

    def possible_blk(self,board,blank_pos,blank_val,b_no):
        pno_other = [i for i in range(9) if not i in b_no]
        p3 = [[] for i in range(9-len(b_no))]
        y3 = []
        row = [[board[y][x] for x in range(9) if board[y][x]!='.'] for y in range(9)]
        col = [[board[y][x] for y in range(9) if board[y][x]!='.'] for x in range(9)]
        for k in range(len(p3)):
            no = pno_other[k]
            p = [0] * len(blank_pos[no])
            for i in range(len(p)):
                y,x = no//3*3 + blank_pos[no][i]//3,(no%3)*3 + blank_pos[no][i]%3
                p[i] = [m for m in blank_val[no] if not m in (row[y]+col[x])]
            y3.append(self.small_permutations(p))
        while True:
            for k in range(len(p3)):
                try:
                    res = next(y3[k])
                    p3[k].append(res)
                except StopIteration as e:
                    return [pno_other[k],p3[k]]

    def small_permutations(self,p,per=[]):
        cur_floor = len(per)
        for pos in range(len(p[cur_floor])):
            t = per + [p[cur_floor][pos]]
            if len(set(t)) == len(t):
                if len(t) == len(p):
                        yield [p[cur_floor][pos]]
                else:
                        for result in self.small_permutations(p,per + [p[cur_floor][pos]]):
                                yield [p[cur_floor][pos]] + result

result = Solution37().solveSudoku([["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]])
for r in result:
    print(r)
print("-------------------------------")

在速度提高上下了功夫：