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| class Solution {
public boolean isValidSudoku(char[][] board) {
Map<Integer, Set<Character>> rowRecorder = new HashMap<>();
Map<Integer, Set<Character>> colRecorder = new HashMap<>();
Map<Integer, Set<Character>> boxRecorder = new HashMap<>();
for (int row = 0; row < board.length; row++) {
for (int col = 0; col < board[0].length; col++) {
if (board[row][col] != '.') {
rowRecorder.putIfAbsent(row, new HashSet<>());
colRecorder.putIfAbsent(col, new HashSet<>());
int boxIndex = (row / 3) * 3 + col / 3;
boxRecorder.putIfAbsent(boxIndex, new HashSet<>());
if (!rowRecorder.get(row).add(board[row][col]) || !colRecorder.get(col).add(board[row][col])
|| !boxRecorder.get(boxIndex).add(board[row][col])) {
return false;
}
}
}
}
return true;
}
}
|
使用hashset的add会返回boolean, 这个很好用.
时间复杂度: O(9 * 9) = O(1)
空间复杂度: O(9 * 9 * 3) = O(1) 同一个非.的字符会被存储三次.
