LeetCode36 Valid Sudoku

题目：

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid. （Easy）

 

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

分析：

就是把行、列和每个九宫格分别判定即可。思路很简单，就是代码需要简化，自己的版本写起来比较复杂，但是感觉好读一些。

代码1：

 1 class Solution {
 2 public:
 3     bool isValidSudoku(vector<vector<char>>& board) {
 4         int flag[10] = {0};
 5         for (int i = 0; i < 9; ++i) {
 6             memset(flag,0,sizeof(flag));
 7             for (int j = 0; j < 9; ++j) {
 8                 if (board[i][j] != '.') {
 9                     if (flag[board[i][j] - '0'] == 1) {
10                         return false;
11                     }
12                     else  {
13                         flag[board[i][j] - '0'] = 1;
14                     }
15                 }
16             }
17         }
18         for (int i = 0; i < 9; ++i) {
19             memset(flag,0,sizeof(flag));
20             for (int j = 0; j < 9; ++j) {
21                 if (board[j][i] != '.') {
22                     if (flag[board[j][i] - '0'] == 1) {
23                         return false;
24                     }
25                     else  {
26                         flag[board[j][i] - '0'] = 1;
27                     }
28                 }
29             }
30         }
31         for (int i = 0; i < 3; ++i) {
32             for (int j = 0; j < 3; ++j) {
33                 memset(flag,0,sizeof(flag));
34                 for (int m = 3 * i; m < 3 * i + 3; ++m) {
35                     for (int n = 3 * j; n < 3 * j + 3; ++n) {
36                         if (board[m][n] != '.') {
37                             if (flag[board[m][n] - '0'] == 1) {
38                                 return false;
39                             }
40                             else  {
41                                 flag[board[m][n] - '0'] = 1;
42                             }
43                         }
44                     }
45                 }
46             }
47         }
48         return true;
49     }
50 };

代码2：

 1 class Solution {
 2 public:
 3     bool isValidSudoku(vector<vector<char>>& board) {
 4         for (int i = 0; i < 9; ++i) {
 5             int flag1[10] = {0};
 6             int flag2[10] = {0};
 7             int flag3[10] = {0};
 8             int m = (i / 3) * 3 , n = (i % 3) * 3;
 9             for (int j = 0; j < 9; ++j) {
10                 if (board[i][j] != '.') {
11                     if (flag1[board[i][j] - '0'] == 1) {
12                         return false;
13                     }
14                     else  {
15                         flag1[board[i][j] - '0'] = 1;
16                     }
17                 }
18                 if (board[j][i] != '.') {
19                     if (flag2[board[j][i] - '0'] == 1) {
20                         return false;
21                     }
22                     else  {
23                         flag2[board[j][i] - '0'] = 1;
24                     }
25                 }
26                 int x = m + (j / 3);
27                 int y = n + (j % 3);
28                 if (board[x][y] != '.') {
29                     if (flag3[board[x][y] - '0'] == 1) {
30                         return false;
31                     }
32                     else  {
33                         flag3[board[x][y] - '0'] = 1;
34                     }
35                 }
36             }
37         }
38         return true;
39     }
40 };

 

 
posted @ 2016-08-24 21:26  wangxiaobao1114  阅读(125)  评论(0)  编辑  收藏  举报