LeetCode刷题（python版）——Topic37解数独
一、题设
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：
- 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。（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：
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：
二、基本思路
本题使用 回溯 （递归）+ 深度遍历算法(dfs) 可较为方便的求解。
1. 首先将board数组转变为数字数组table运算较为简便，等到所有对于矩阵table的运算结束之后再转变成字符数组board。
2. 其次从(x = 0,y = 0)处开始dfs，分别产生以下几种情况：
2.1 当x = 9时，即为第十行，则意味着全部的遍历填数早已结束，则返回True
2.2 当y = 9时，即为每行的第十列，则意味着每行到达了尽头，应该换行了，则递归返回下一行第一个数：dfs(x+1,0)
2.3 当处理后的table[x][y]不为0时，即代表该坐标(x,y)在数独中是初始值，则跳过看下一个位置是否为空：dfs(x,y+1)
2.4 当以上全部的三种情况处理好之后，可以开始填入数据了，则从0~9挨个尝试填入，若当前数val不满足：行唯一、列唯一以及九宫格内唯一这三个条件时，则将当前table[x][y]重置为0，尝试下个数val+1；若三个条件同时满足，既可以填入val并当所有的dfs(x,y+1)成立时，这次填入的所有数全部成立，返回True。
三、代码实现
def solveSudoku(self, board):
# tag1将board先转成数字数组方便存放
table = [([0] * 9) for i in range(9)]
# dfs func start……
def dfs(x,y):
# 如果x是第十行（table已全部遍历结束）
if x == 9:
return True
# 如果y是9，则该换行了
if y == 9:
return dfs(x+1,0)
# 如果(x,y)有数字，则跳转到下一个格子
if table[x][y] != 0:
return dfs(x,y+1)
# val代表预计填入0~9的数字
for val in range(10):
# 看是否满足三个条件，有一个不满足标志flag记为false
flag = True
# 行是否唯一
for col in range(9):
if table[x][col] == val:
flag = False
break
# 列是否唯一
for row in range(9):
if table[row][y] == val:
flag = False
break
# (xx,yy)代表每个 3 * 3 方格起点坐标
# 九宫格内是否唯一
xx,yy = x / 3 * 3 , y / 3 * 3
for i in range(3):
for j in range(3):
if table[xx + i][yy + j] == val:
flag = False
break
# 若唯一
if flag == True:
table[x][y] = val #val可填入
if dfs(x,y+1):#全部填入成功
return True
table[x][y] = 0 #当前val填入失败，尝试下一个val
# dfs func end……
# main func start……
for i in range(9):
for j in range(9):
if board[i][j] != '.':
table[i][j] = int(board[i][j])
dfs(0,0)
# tag1将table先转成字符数组board返回
for i in range(9):
for j in range(9):
board[i][j] = str(table[i][j])
# main func end……四、效率总结
