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 ‘.‘.
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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