正文
Java for LeetCode 036 Valid Sudoku
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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 '.'.
解题思路：
传说中的数独(九宫格)问题，老实遍历三个规则即可：
JAVA实现：
static public boolean isValidSudoku(char[][] board) {
for (int i = 0; i < board.length; i++) {
HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>();
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] != '.') {
if (hashmap.containsKey(board[i][j]))
return false;
hashmap.put(board[i][j], 1);
}
}
}
for (int j = 0; j < board[0].length; j++) {
HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>();
for (int i = 0; i < board.length; i++) {
if (board[i][j] != '.') {
if (hashmap.containsKey(board[i][j]))
return false;
hashmap.put(board[i][j], 1);
}
}
}
for (int i = 0; i < board.length; i += 3){
for (int j = 0; j < board[0].length; j += 3){
HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>();
for (int k = 0; k < 9; k++) {
if (board[i + k / 3][j + k % 3] != '.') {
if (hashmap.containsKey(board[i + k / 3][j + k % 3]))
return false;
hashmap.put(board[i + k / 3][j + k % 3], 1);
}
}
}
}
return true;
}
