title: LeetCode No.37
categories:
- OJ
- LeetCode
tags:
- Programing
- LeetCode
- OJ
LeetCode第三十七题
今天三十了，新年快乐哈。
题目描述
编写一个程序，通过填充空格来解决数独问题。
一个数独的解法需遵循如下规则：
数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
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。
一个数独。
答案被标成红色。
提示：
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。
回撤法
回撤法标准框架:
def backtrack(path, selected):
if 满足停止条件：
res.append(path)
for 选择 in 选择列表：
做出选择
递归执行backtrack
满足则return True
如果不满足要求就撤销选择
代码
class Solution(object):
# 检查选择要填充的数字是否在一行/一列/九个格子里面出现
def check(self, board, row, col, c):
for i in range(9):
if board[row][i] == c:
return False
if board[i][col] == c:
return False
if board[(row//3)*3 + i // 3][(col//3)*3 + i % 3] == c:
return False
return True
# 回溯法
def backtrack(self,board,i,j):
# 停止条件
## 一行访问完之后跳转到下一行
if j == 9:
return self.backtrack(board,i+1,0)
## 所有行访问完之后，返回True
if i == 9:
return True
## 如果不为.，则不用管,继续访问下一列
if board[i][j] != '.':
return self.backtrack(board,i,j+1)
# 选择操作
## 遍历选择列表，此处的选择是，给空白处填 "1" - "9" 中之一
for k in range(1,10):
c = str(k)
if not self.check(board, i, j, c): # 判断选择的字符是否满足要求（不与其他位置冲突）
continue
board[i][j] = c # 做出选择
if self.backtrack(board, i, j + 1): # 递归调用，直接return是因为只需要一个可行解，而不需要所有可行解
return True
board[i][j] = '.' # 撤销选择
def solveSudoku(self, board):
"""
:type board: List[List[str]]
:rtype: None Do not return anything, modify board in-place instead.
核心思想：采用回溯法
回溯法的标准框架
def backtrack(path, selected):
if 满足停止条件：
res.append(path)
for 选择 in 选择列表：
做出选择
递归执行backtrack
满足则return True
如果不满足要求就撤销选择
"""
self.backtrack(board,0,0)
return board
if __name__ == '__main__':
s = Solution()
print(s.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"]]))