linear-list/array/valid-sudoku
Valid Sudoku
描述
Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules http://sudoku.com.au/TheRules.aspx .
The Sudoku board could be partially filled, where empty cells are filled with the character '.'.
分析
细节实现题。
代码
// Valid Sudoku
// 时间复杂度O(n^2)，空间复杂度O(1)
public class Solution {
public boolean isValidSudoku(char[][] board) {
boolean[] used = new boolean[9];
for (int i = 0; i < 9; ++i) {
Arrays.fill(used, false);
for (int j = 0; j < 9; ++j) // 检查行
if (!check(board[i][j], used))
return false;
Arrays.fill(used, false);
for (int j = 0; j < 9; ++j) // 检查列
if (!check(board[j][i], used))
return false;
}
for (int r = 0; r < 3; ++r) // 检查 9 个子格子
for (int c = 0; c < 3; ++c) {
Arrays.fill(used, false);
for (int i = r * 3; i < r * 3 + 3; ++i)
for (int j = c * 3; j < c * 3 + 3; ++j)
if (!check(board[i][j], used))
return false;
}
return true;
}
private static boolean check(char ch, boolean[] used) {
if (ch == '.') return true;
if (used[ch - '1']) return false;
return used[ch - '1'] = true;
}
};
// Valid Sudoku
// 时间复杂度O(n^2)，空间复杂度O(1)
class Solution {
public:
bool isValidSudoku(const vector<vector<char>>& board) {
bool used[9];
for (int i = 0; i < 9; ++i) {
fill(used, used + 9, false);
for (int j = 0; j < 9; ++j) // 检查行
if (!check(board[i][j], used))
return false;
fill(used, used + 9, false);
for (int j = 0; j < 9; ++j) // 检查列
if (!check(board[j][i], used))
return false;
}
for (int r = 0; r < 3; ++r) // 检查 9 个子格子
for (int c = 0; c < 3; ++c) {
fill(used, used + 9, false);
for (int i = r * 3; i < r * 3 + 3; ++i)
for (int j = c * 3; j < c * 3 + 3; ++j)
if (!check(board[i][j], used))
return false;
}
return true;
}
bool check(char ch, bool used[9]) {
if (ch == '.') return true;
if (used[ch - '1']) return false;
return used[ch - '1'] = true;
}
};