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Python描述 LeetCode 37. 解数独


Python描述 LeetCode 37. 解数独

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Python描述 LeetCode 37. 解数独_数独

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题目

编写一个程序,通过填充空格来解决数独问题。

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。(请参考示例图)

数独部分空格内已填入了数字,空白格用 ​​'.'​​ 表示。

示例 1:

Python描述 LeetCode 37. 解数独_数独_02

输入:board = [["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"]]
输出:[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释:输入的数独如上图所示,唯一有效的解决方案如下所示:

提示:

  • ​board.length == 9​
  • ​board[i].length == 9​
  • ​board[i][j]​​​ 是一位数字或者​​'.'​
  • 题目数据保证输入数独仅有一个解

解题思路

DFS一路向前搜索,挨个将空格填上数字,当全部满足时,返回结束循环

Python描述

class Solution:
def solveSudoku(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
global complete_flag
line = [[0 for _ in range(10)] for __ in range(9)]
col = [[0 for _ in range(10)] for __ in range(9)]
jiu = [[[0 for _ in range(10)] for __ in range(3)] for ___ in range(3)]
complete_flag = False

# 初始化
for i in range(9):
for j in range(9):
tmp = int(board[i][j]) if board[i][j] != '.' else 0
if tmp == 0:
continue
line[i][tmp] += 1
col[j][tmp] += 1
jiu[i // 3][j // 3][tmp] += 1

def whether_fit(i, j, num):
# 检测当前位置是否能放num
return line[i][num] + col[j][num] + jiu[i // 3][j // 3][num] == 0

def dfs(i, j):
global complete_flag
if complete_flag:
return
if i == j == 8 and board[i][j] != '.':
complete_flag = True
return
if board[i][j] != '.':
n_j = (j + 1) % 9
n_i = i if n_j != 0 else i + 1
dfs(n_i, n_j)
return
for ch in "123456789":
tmp = int(ch)
if whether_fit(i, j, tmp):
board[i][j] = ch
line[i][tmp] += 1
col[j][tmp] += 1
jiu[i // 3][j // 3][tmp] += 1

n_j = (j + 1) % 9
n_i = i if n_j != 0 else i + 1
if n_i == 9:
complete_flag = True
return
dfs(n_i, n_j)
if complete_flag:
return
board[i][j] = "."
line[i][tmp] -= 1
col[j][tmp] -= 1
jiu[i // 3][j // 3][tmp] -= 1

dfs(0, 0)


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