蜗牛慢慢爬 LeetCode 36.Valid Sudoku [Difficulty: Medium]

题目

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 '.'.

蜗牛慢慢爬 LeetCode 36.Valid Sudoku [Difficulty: Medium]

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.

翻译

数独嘛 略

Hints

Related Topics: Hash Table

通过哈希表实现O(n^2)的算法 即只遍历整个数独的表一遍就可以得到有效性判断

代码

C++

class Solution
{
public:
bool isValidSudoku(vector<vector<char> > &board)
{
int used1[9][9] = {0}, used2[9][9] = {0}, used3[9][9] = {0}; for(int i = 0; i < board.size(); ++ i)
for(int j = 0; j < board[i].size(); ++ j)
if(board[i][j] != '.')
{
int num = board[i][j] - '0' - 1, k = i / 3 * 3 + j / 3;
if(used1[i][num] || used2[j][num] || used3[k][num])
return false;
used1[i][num] = used2[j][num] = used3[k][num] = 1;
} return true;
}
};

Java

//an interesting solution from discuss
public boolean isValidSudoku(char[][] board) {
Set seen = new HashSet();
for (int i=0; i<9; ++i) {
for (int j=0; j<9; ++j) {
char number = board[i][j];
if (number != '.')
if (!seen.add(number + " in row " + i) ||
!seen.add(number + " in column " + j) ||
!seen.add(number + " in block " + i/3 + "-" + j/3))
return false;
}
}
return true;
}

Python

class Solution(object):
def isValidSudoku(self, board):
"""
:type board: List[List[str]]
:rtype: bool
"""
row = {}
col = {}
cube = {}
for i in range(0,9):
row[i] = set()
for j in range(0,9):
if j not in col: col[j] = set()
if board[i][j] !='.':
num = int(board[i][j])
k = i/3*3+j/3
if k not in cube: cube[k] = set()
if num in row[i] or num in col[j] or num in cube[k]:
return False
row[i].add(num)
col[j].add(num)
cube[k].add(num)
return True
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