From : https://leetcode.com/problems/valid-sudoku/
Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

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 '.'.
只要不重复出现即可。
A partially filled sudoku which is valid.
class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
// **
set<int> box;
char ch;
for(int i=0; i<9; i++) {
box.clear();
for(int j=0; j<9; j++) {
ch = board[i][j];
if(ch != '.' && box.find(ch)!=box.end()) {
return false;
} else if(ch != '.') {
box.insert(ch);
}
}
}
// **
for(int j=0; j<9; j++) {
box.clear();
for(int i=0; i<9; i++) {
ch = board[i][j];
if(ch != '.' && box.find(ch)!=box.end()) {
return false;
} else if(ch != '.') {
box.insert(ch);
}
}
}
// **
for(int i=0; i<3; i++) {
for(int j=0; j<3; j++) {
box.clear();
for(int k=i*3; k<(i+1)*3;k++) {
for(int t=j*3; t<(j+1)*3;t++) {
ch = board[k][t];
if(ch != '.' && box.find(ch)!=box.end()) {
return false;
} else if(ch != '.') {
box.insert(ch);
}
}
}
}
}
return true;
}
};
class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
vector<vector<bool>> rows(9, vector<bool>(9,false));
vector<vector<bool>> cols(9, vector<bool>(9,false));
vector<vector<bool>> blocks(9, vector<bool>(9,false));
for(int i = 0; i < 9; i++)
for(int j = 0; j < 9; j++) {
if(board[i][j] == '.')continue;
int num = board[i][j] - '1';
if(rows[i][num] || cols[j][num] || blocks[i - i%3 + j/3][num])
return false;
rows[i][num] = cols[j][num] = blocks[i - i%3 + j/3][num] = true;
}
return true;
}
};
public class Solution {
public boolean isValidSudoku(char[][] board) {
boolean[][] rows = new boolean[9][9];
boolean[][] cols = new boolean[9][9];
boolean[][] blocks = new boolean[9][9];
int dot = '.' - '1';
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
int n = board[i][j] - '1';
if (dot == n) {
continue;
}
if (rows[i][n] || cols[j][n] || blocks[i / 3 * 3 + j / 3][n]) {
return false;
}
rows[i][n] = true;
cols[j][n] = true;
blocks[i / 3 * 3 + j / 3][n] = true;
}
}
return true;
}
}