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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.
分析：检验数独是否有效。只用检查已经填充的部分，没填充的部分用 ‘.‘ 表示。
分别检查行、列以及小九宫格。为了避免麻烦，我用了map来迅速查找数字。运行时间46ms。
1 class Solution { 2 public: 3 bool isValidSudoku(vector<vector<char>>& board) { 4 for(int i = 0; i < 9; i++){ 5 //check each row 6 map<char, int> m; 7 for(int j = 0; j < 9; j++){ 8 if(board[i][j] != ‘.‘){ 9 if(m.find(board[i][j]) != m.end()) return false; 10 else m[board[i][j]] = 1; 11 } 12 } 13 14 //check each column 15 map<char, int> n; 16 for(int k = 0; k < 9; k++){ 17 if(board[k][i] != ‘.‘){ 18 if(n.find(board[k][i]) != n.end()) return false; 19 else n[board[k][i]] = 1; 20 } 21 } 22 } 23 //check each sub-boxes 24 for(int i = 0; i < 9; i += 3){ 25 for(int j = 0; j < 9; j += 3){ 26 map<char, int> mm; 27 for(int p = i; p < i + 3; p++){ 28 for(int q = j; q < j + 3; q++){ 29 if(board[p][q] != ‘.‘){ 30 if(mm.find(board[p][q]) != mm.end()) return false; 31 else{ 32 mm[board[p][q]] = 1; 33 } 34 } 35 } 36 } 37 } 38 } 39 return true; 40 } 41 };
为了进一步优化，可以不用map进行数字重复测试。通过使用一个bool型数组，每出现一个数字，便将该数字相应下标置1，运行时间16ms！
1 class Solution { 2 public: 3 bool isValidSudoku(vector<vector<char>>& board) { 4 for(int i = 0; i < 9; i++){ 5 //check each row 6 bool arrRow[9]; 7 fill(arrRow, arrRow + 9, false); 8 for(int j = 0; j < 9; j++){ 9 if(board[i][j] != ‘.‘){ 10 if(arrRow[board[i][j] - ‘1‘]) return false; 11 else arrRow[board[i][j] - ‘1‘] = true; 12 } 13 } 14 15 //check each column 16 bool arrCol[9]; 17 fill(arrCol, arrCol + 9, false); 18 for(int k = 0; k < 9; k++){ 19 if(board[k][i] != ‘.‘){ 20 if(arrCol[board[k][i] - ‘1‘]) return false; 21 else arrCol[board[k][i] - ‘1‘] = true; 22 } 23 } 24 } 25 //check each sub-boxes 26 for(int i = 0; i < 9; i += 3){ 27 for(int j = 0; j < 9; j += 3){ 28 bool arr[9]; 29 fill(arr, arr + 9, false); 30 for(int p = i; p < i + 3; p++){ 31 for(int q = j; q < j + 3; q++){ 32 if(board[p][q] != ‘.‘){ 33 if(arr[board[p][q] - ‘1‘]) return false; 34 else arr[board[p][q] - ‘1‘] = true; 35 } 36 } 37 } 38 } 39 } 40 return true; 41 } 42 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4500985.html
