LeetCode37 Sudoku Solver

题目：

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character ‘.‘.

You may assume that there will be only one unique solution.

A sudoku puzzle...

...and its solution numbers marked in red.  （Hard）

分析：

DFS搜索，但是不太好写。依次尝试所有可能的数字，然后递归地继续向后添加，发现当填入任何数字都无法满足条件（行列九宫格）后返回false（说明前面填错了）。

想清楚递归地过程的话也不是很难理解。

注意事项：

1.每个数字添加后判断本行，本列和所在九宫格即可，不必判断所有数独内容；

2.自己依照上一题isValid函数写的条件判定代码。导致少了一个board[i][j] != ‘.‘的判断（见代码注释部分），导致所有情况都返回false（因为其会对flag[‘.‘ - ‘0‘]赋值为1，然后判断等于1）这里错了好久才发现。

代码：

 1 class Solution {
 2 private:
 3     bool isValid (vector<vector<char>>& board, int x, int y) {
 4         int flag[10] = {0};
 5         for (int i = 0; i < board.size(); ++i) {
 6             if (board[x][i] != ‘.‘) {                            //少了这句会导致这个判断总是false!!!
 7                 if (flag[board[x][i] - ‘0‘] == 1) {
 8                     return false;
 9                 }
10                 else {
11                     flag[board[x][i] - ‘0‘] = 1;
12                 }
13             }
14         }
15         memset(flag,0,sizeof(flag));
16         for (int i = 0; i < board.size(); ++i) {
17             if (board[i][y] != ‘.‘) {
18                 if (flag[board[i][y] - ‘0‘] == 1) {
19                     return false;
20                 }
21                 else {
22                     flag[board[i][y] - ‘0‘] = 1;
23                 }
24             }
25         }
26         memset(flag,0,sizeof(flag));
27         int sqx = (x / 3) * 3;
28         int sqy = (y / 3) * 3;
29         for (int i = sqx; i < sqx + 3; ++i) {
30             for (int j = sqy; j < sqy + 3; ++j) {
31                 if (board[i][j] != ‘.‘) {
32                     if (flag[board[i][j] - ‘0‘] == 1) {
33                         return false;
34                     }
35                     else {
36                         flag[board[i][j] - ‘0‘] = 1;
37                     }
38                 }
39             }
40         }
41         return true;
42     }
43     bool solvehelper(vector<vector<char>>& board) {
44         for (int i = 0; i < board.size(); ++i) {
45             for (int j = 0; j < board.size(); ++j) {
46                 if (board[i][j] == ‘.‘) {
47                     for (int k = 0; k < 9; ++k)  {
48                         board[i][j] = ‘1‘ + k;
49                         if (isValid(board, i, j)) {
50                             if (solvehelper(board)) {
51                                 return true;
52                             }
53                         }
54                     }
55                     board[i][j] = ‘.‘;
56                     return false;
57                 }
58             }
59         }
60         return true;
61     }
62 public:
63     void solveSudoku(vector<vector<char>>& board) {
64         bool b = solvehelper(board);
65     }
66 };