[抄题]:

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

  1. Each row must contain the digits 1-9 without repetition.
  2. Each column must contain the digits 1-9 without repetition.
  3. Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.


A partially filled sudoku which is valid.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'

[暴力解法]:

时间分析:

空间分析:

[优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

[思维问题]:

不知道怎么判断每个单独的3*3: 三倍行数遍历+一倍列数遍历进入具体的3*3

[英文数据结构或算法,为什么不用别的数据结构或算法]:

[一句话思路]:

每一行都新建三个集合,然后行重复为[i][j] 列重复为[j][i]

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[五分钟肉眼debug的结果]:

[总结]:

三倍行数遍历+一倍列数遍历进入具体的3*3

[复杂度]:Time complexity: O(mn) Space complexity: O(mn)

[算法思想:迭代/递归/分治/贪心]:

[关键模板化代码]:

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

[代码风格] :

[是否头一次写此类driver funcion的代码] :

[潜台词] :

class Solution {
public boolean isValidSudoku(char[][] board) { //for loop, judge row, col, and box
for (int i = 0; i < 9; i++) {
//initialization: 3 sets in each new row
HashSet<Character> row = new HashSet<>();
HashSet<Character> col = new HashSet<>();
HashSet<Character> box = new HashSet<>(); for (int j = 0; j < 9; j++) {
//judge row, col
if (board[i][j] != '.' && !row.add(board[i][j])) return false;
if (board[j][i] != '.' && !col.add(board[j][i])) return false;
//judge box
//same for each row
int rowIndex = 3 * (i / 3);
System.out.println("rowIndex = " + rowIndex);
int colIndex = 3 * (i % 3);
System.out.println("colIndex = " + colIndex);
System.out.println("-------------");
//different for each col
if (board[rowIndex + j / 3][colIndex + j % 3] != '.' && !box.add(board[rowIndex + j / 3][colIndex + j % 3])) return false;
}
} return true;
}
}

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