Description

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.

The original problem is here.

The original code is here.

My Solution

I solve this problem in C++, as below:

``````/*
*Valid Sudoku
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<stdlib.h>
using namespace std;

class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
for(int row=0;row<board.size();row++){
if(!isValid(board,row,row,0,board[row].size()-1))
return false;
}
for(int col=0;col<board[0].size();col++){
if(!isValid(board,0,board.size()-1,col,col))
return false;
}
for(int row=0;row<board.size();row+=3){
for(int col=0;col<board[0].size();col+=3){
if(!isValid(board,row,row+2,col,col+2))
return false;
}
}
return true;
}
bool isValid(vector<vector<char>>& board, int rowi, int rowj, int coli, int colj) {
int index[10]={0};
int num;
for(int row=rowi;row<=rowj;row++){
for(int col=coli;col<=colj;col++){
if(board[row][col]!='.'){
num = board[row][col] - '0';
index[num] += 1;
if(index[num] > 1)
return false;
}
}
}
return true;
}
};
``````

Note

To solve the problem, just check whether each row(each column, each 3*3 area) has repeat digitals resoectively.