題目如下：
當我們拿着數獨遊戲時，我們是通過觀感直覺來求解的。詳細說來：
- 我們會去找那些規則下能夠確定唯一解的空格，然後填入唯一解後再去尋找其他可以確定唯一解的空格。
- 如果找不到有唯一解，我們就找解空間比較小的空格，從解空間中挑一個填上去試試，如果可行繼續嘗試步驟1，2。如果不可行（有一些空格沒有解），我們就把嘗試的解從解空間中去掉，再挑其他解進行嘗試。
我採用的解題思路如下，爲了方便理解，代碼有點長，但用斷點跟一遍很容易懂，代碼 beat 80%：
- 通過數獨規則確定每行/ 每列/ 每3*3 box 的候選解空間，如果解空間都爲空，代表數獨解出，返回答案。反之，步驟2。
- 對於每個空格基於（1）中對應解空間的交集確定解空間，如果有空格解空間爲空，返回錯誤。
- 找出那些有唯一解的空格，如果有填入，再重複步驟（遞歸）
- 如果沒有唯一解的空格，挑選解空間最小的空格，挑選空間中的一個解填入，然後重複步驟（遞歸），如果返回錯誤，就將該解從解空間中刪除，如果刪除後解空間爲空，返回錯誤。反之，再挑一個解填入，重複步驟（遞歸）。
算法參考流程圖如下：
參考代碼如下：
class Solution:
def solver(self, board):
import copy
# get the candidates based on rule of row
cd_row = [["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8", "9"],
["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8", "9"],
["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8", "9"]]
i = 0
while i < 9:
for n in board[i]:
if n in cd_row[i]:
cd_row[i].remove(n)
i += 1
# Terminal condition
T = 0
for r in cd_row:
if len(r) == 0:
T = T + 1
if T == 9:
return board
# get the candidates based on rule of col
cd_col = [["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8", "9"],
["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8", "9"],
["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8", "9"]]
j = 0
while j < 9:
for row in board:
if row[j] in cd_col[j]:
cd_col[j].remove(row[j])
j += 1
# get candidates based on rule of 3*3 box
cd_box = [["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8", "9"],
["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8", "9"],
["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8",
"9"], ["1", "2", "3", "4", "5", "6", "7", "8", "9"]]
s = 0
b = 0
while s <= 6:
col = 0
while col <= 6:
for row in board[s:s + 3]:
for n in row[col:col + 3]:
if n in cd_box[b]:
cd_box[b].remove(n)
b += 1
col += 3
s += 3
min_cd = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]
p_x = 0 # solve point x-th row
p_y = 0 # solve point y-th col
cert = [] # point only has one candidate
a = 0
cd_board = []
while a < 9:
cd_b = []
b = 0
while b < 9:
if board[a][b] != ".":
cd_b.append(board[a][b])
else:
cd = [
c for c in cd_row[a] if c in cd_col[b]
and c in cd_box[(a // 3) * 3 + b // 3]
]
if len(cd) == 0:
cert = []
return "wrong"
elif len(cd) <= len(min_cd):
min_cd = cd
p_x = a
p_y = b
if len(cd) == 1:
cert.append([p_x, p_y])
cd_b.append(cd)
b += 1
cd_board.append(cd_b)
a += 1
# print(cd_board, cert)
if len(cert) > 0:
for p in cert:
temp = copy.deepcopy(board)
temp[p[0]][p[1]] = cd_board[p[0]][p[1]][0]
# print("board:", board)
res = self.solver(temp)
if res == "wrong":
return "wrong"
else:
return res
else:
trynum = min_cd[0]
temp = copy.deepcopy(board)
temp[p_x][p_y] = trynum
res = self.solver(temp)
while res == "wrong":
if len(min_cd) > 1:
min_cd.remove(trynum)
trynum = min_cd[0]
temp[p_x][p_y] = trynum
res = self.solver(temp)
else:
return "wrong"
return res
def solveSudoku(self, board):
res = self.solver(board)
i = 0
if res != "wrong":
while i < 9:
j = 0
while j < 9:
board[i][j] = res[i][j]
j = j + 1
i = i + 1
注意：
- 深拷貝和淺拷貝的問題。
- 要求直接對 board進行修改，不可以返回值。
源碼地址：
https://github.com/jediL/LeetCodeByPython
其它題目：[leetcode題目答案講解彙總（Python版 持續更新）]
(https://www.jianshu.com/p/60b5241ca28e)
ps：如果您有好的建議，歡迎交流 :-D，
也歡迎訪問我的個人博客 苔原帶 (www.tundrazone.com)