Sudoku Solver [LeetCode]

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.

Summary: finds the cell which has the least candidates first, then checks all candidates in the cell.

 1    vector<int> findLeastCandidate(vector<vector<char> > &board, vector<int> &out_digits){
 2         vector<int> idx;
 3         int candidate_num = 10;
 4         int n = board.size();
 5         for(int i = 0; i < n; i++){
 6             for(int j = 0; j < n; j ++) {
 7                 if(board[i][j] == '.'){
 8                     static const int arr[] = {1,2,3,4,5,6,7,8,9};
 9                     vector<int> digits(arr, arr + sizeof(arr) / sizeof(arr[0]));
10                     //check row
11                     for(int k = 0; k < n; k ++){
12                         if(board[i][k] != '.')
13                             digits[board[i][k] - 48 - 1] = -1;
14                     }
15                     //check column
16                     for(int k = 0; k < n; k ++){
17                         if(board[k][j] != '.')
18                             digits[board[k][j] - 48 - 1] = -1;
19                     }
20                     //check the 9 sub-box
21                     int row_box_idx = i/3;
22                     int column_box_idx = j/3;
23                     for(int l = row_box_idx * 3; l < (row_box_idx + 1) * 3; l ++){
24                         if(l == i)
25                             continue;
26                         for(int m = column_box_idx * 3; m < (column_box_idx + 1) * 3; m ++){
27                             if(board[l][m] != '.' && m != j)
28                                 digits[board[l][m] - 48 - 1] = -1;
29                         }
30                     }
31                     //caculate candidate number
32                     int tmp_num = 0;
33                     for(int candidate: digits)
34                         if(candidate != -1) 
35                             tmp_num ++;
36     
37                     if(tmp_num == 0){
38                         if(idx.size() == 0){
39                             idx.push_back(-1);
40                             idx.push_back(-1);
41                         }else {
42                             idx[0] = -1;
43                             idx[1] = -1;
44                         }
45                         return idx;
46                     }
47     
48                     if(tmp_num < candidate_num){
49                         candidate_num = tmp_num;
50                         out_digits = digits;
51                         if(idx.size() == 0){
52                             idx.push_back(i);
53                             idx.push_back(j);
54                         }else {
55                             idx[0] = i;
56                             idx[1] = j;
57                         }
58                     }
59                 }
60             }
61         }
62         return idx;
63     }
64 
65     bool isValidSudoku(vector<vector<char> > &board) {
66         //find the candidate which has most constrict
67         vector<int> digits;
68         vector<int> idx = findLeastCandidate(board, digits);
69         if(idx.size() == 0)
70             return true;
71     
72         if(idx[0] == -1)
73             return false;
74     
75         int i = idx[0];
76         int j = idx[1];
77         //recursive
78         bool is_all_minus = true;
79         for(int candidate: digits) {
80             if(candidate != -1) {
81                 is_all_minus = false;
82                 board[i][j] = candidate + 48;
83                 if(isValidSudoku(board))
84                     return true;
85                 else 
86                     board[i][j] = '.';
87             }
88         }
89         if(is_all_minus)
90             return false;
91         return false;
92     }
93     
94     void solveSudoku(vector<vector<char> > &board) {
95         isValidSudoku(board);
96     }

 

posted @ 2013-11-07 12:52  假日笛声  阅读(597)  评论(0)  编辑  收藏  举报