Valid Sudoku
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.
解题思路:
题意为验证数独的有效性。这里说的有效是值谜面的有效性,不包括是否能够解出。数独的规则是,每一行1-9只出现一次,每一列1-9只出现一次,每一个小九宫格1-9只出现一次。依次验证即可。一个陷阱就是字符减的不是‘0‘,而是‘1‘
class Solution { public: bool isValidSudoku(vector<vector<char>>& board) { //验证每一行是否有效 for(int i=0; i<9; i++){ if(!checkRowValid(board, i)){ return false; } } //验证每一列是否有效 for(int i=0; i<9; i++){ if(!checkColumnValid(board, i)){ return false; } } //验证每一格是否有效 for(int i=0; i<9; i=i+3){ for(int j=0; j<9; j=j+3){ if(!checkGridValid(board, i, j)){ return false; } } } return true; } //验证每个格是否有效,传入的是左上角的下标 bool checkGridValid(vector<vector<char>>& board, int m, int n){ bool flag[9]; memset(flag, 0, sizeof(bool)*9); for(int i=m; i<m+3; i++){ for(int j=n; j<n+3; j++){ if(board[i][j]=='.'){ continue; } if(flag[board[i][j]-'1']){ return false; } flag[board[i][j]-'1']=true; } } return true; } //验证每一行是否有效,传入的是行号 bool checkRowValid(vector<vector<char>>& board, int m){ bool flag[9]; memset(flag, 0, sizeof(bool)*9); for(int i=0; i<9; i++){ if(board[m][i]=='.'){ continue; } if(flag[board[m][i]-'1']){ return false; } flag[board[m][i]-'1']=true; } return true; } //验证每一列是否有效,传入的是列号 bool checkColumnValid(vector<vector<char>>& board, int n){ bool flag[9]; memset(flag, 0, sizeof(bool)*9); for(int i=0; i<9; i++){ if(board[i][n]=='.'){ continue; } if(flag[board[i][n]-'1']){ return false; } flag[board[i][n]-'1']=true; } return true; } };
原文地址:http://blog.csdn.net/kangrydotnet/article/details/45419481