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题目要求:
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.
Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
这道题解法暂没有比较巧妙的,所以下边程序所用方法就是Brute Force(暴力破解):
1 class Solution { 2 public: 3 int charToInt(char c) 4 { 5 char str[] = {‘1‘, ‘2‘, ‘3‘, ‘4‘, ‘5‘, ‘6‘, ‘7‘, ‘8‘, ‘9‘}; 6 int intStr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; 7 for(int i = 0; i < 9; i++) 8 { 9 if(c == str[i]) 10 return intStr[i]; 11 } 12 return -1; 13 } 14 15 bool isValidSudoku(vector<vector<char>>& board) { 16 unordered_map<int, int> hashMap; 17 for(int i = 1; i < 10; i++) 18 hashMap[i] = 0; 19 // row by row 20 for(int i = 0; i < 9; i++) 21 { 22 for(int k = 0; k < 10; k++) 23 hashMap[k] = 0; 24 for(int j = 0; j < 9; j++) 25 { 26 int tmp = charToInt(board[i][j]); 27 if(tmp != -1) 28 { 29 hashMap[tmp]++; 30 if(hashMap[tmp] > 1) 31 return false; 32 } 33 } 34 } 35 // column by column 36 for(int j = 0; j < 9; j++) 37 { 38 for(int k = 1; k < 10; k++) 39 hashMap[k] = 0; 40 for(int i = 0; i < 9; i++) 41 { 42 int tmp = charToInt(board[i][j]); 43 if(tmp != -1) 44 { 45 hashMap[tmp]++; 46 if(hashMap[tmp] > 1) 47 return false; 48 } 49 } 50 } 51 // 3*3 boxes by 3*3 boxes 52 for(int i = 0; i < 9; i += 3) 53 { 54 for(int j = 0; j < 9; j += 3) 55 { 56 for(int k = 0; k < 10; k++) 57 hashMap[k] = 0; 58 for(int m = i; m < i + 3; m++) 59 for(int n = j; n < j + 3; n++) 60 { 61 int tmp = charToInt(board[m][n]); 62 if(tmp != -1) 63 { 64 hashMap[tmp]++; 65 if(hashMap[tmp] > 1) 66 return false; 67 } 68 } 69 } 70 } 71 72 return true; 73 } 74 };
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原文地址:http://www.cnblogs.com/xiehongfeng100/p/4590691.html