算法leetcode｜36. 有效的数独（rust重拳出击）

Posted 2023-03-03 二当家的白帽子

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文章目录

• 36. 有效的数独：
• • 样例 1：
  • 样例 2：
  • 提示：
• 分析：
• 题解：
• • rust
  • go
  • c++
  • c
  • python
  • java

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36. 有效的数独：

请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ，验证已经填入的数字是否有效即可。

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）

注意：

• 一个有效的数独（部分已被填充）不一定是可解的。
• 只需要根据以上规则，验证已经填入的数字是否有效即可。
• 空白格用 '.' 表示。

样例 1：

输入：
	board = 
	[["5","3",".",".","7",".",".",".","."]
	,["6",".",".","1","9","5",".",".","."]
	,[".","9","8",".",".",".",".","6","."]
	,["8",".",".",".","6",".",".",".","3"]
	,["4",".",".","8",".","3",".",".","1"]
	,["7",".",".",".","2",".",".",".","6"]
	,[".","6",".",".",".",".","2","8","."]
	,[".",".",".","4","1","9",".",".","5"]
	,[".",".",".",".","8",".",".","7","9"]]
	
输出：
	true

样例 2：

输入：
	board = 
	[["8","3",".",".","7",".",".",".","."]
	,["6",".",".","1","9","5",".",".","."]
	,[".","9","8",".",".",".",".","6","."]
	,["8",".",".",".","6",".",".",".","3"]
	,["4",".",".","8",".","3",".",".","1"]
	,["7",".",".",".","2",".",".",".","6"]
	,[".","6",".",".",".",".","2","8","."]
	,[".",".",".","4","1","9",".",".","5"]
	,[".",".",".",".","8",".",".","7","9"]]
	
输出：
	false
	
解释：
	除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

提示：

• board.length == 9
• board[i].length == 9
• board[i][j] 是一位数字（1-9）或者 ‘.’

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分析：

• 面对这道算法题目，二当家的陷入了沉思。
• 主要是如何存储行，列，以及3*3宫内出现过的值。
• 方法很多，集合，整形数组，布尔数组都可以，只有1-9，一共9个数，最优化的空间方式应该是仅仅用一个整形，然后用位运算。

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题解：

rust

impl Solution 
    pub fn is_valid_sudoku(board: Vec<Vec<char>>) -> bool 
        let mut rows = vec![vec![false; 9]; 9];
        let mut columns = vec![vec![false; 9]; 9];
        let mut sub_boxes = vec![vec![vec![false; 9]; 3]; 3];

        for i in 0..9 
            for j in 0..9 
                let c = board[i][j];
                if c != '.' 
                    let index = (c as u8 - b'1') as usize;

                    if rows[i][index] || columns[j][index] || sub_boxes[i / 3][j / 3][index] 
                        return false;
                    

                    rows[i][index] = true;
                    columns[j][index] = true;
                    sub_boxes[i / 3][j / 3][index] = true;
                
            
        

        return true;
    

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go

func isValidSudoku(board [][]byte) bool 
    var rows, columns [9][9]bool
	var subBoxes [3][3][9]bool

	for i, row := range board 
		for j, c := range row 
			if c != '.' 
				index := c - '1'

				if rows[i][index] || columns[j][index] || subBoxes[i/3][j/3][index] 
					return false
				

				rows[i][index] = true
				columns[j][index] = true
				subBoxes[i/3][j/3][index] = true

			
		
	

	return true

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c++

class Solution 
public:
    bool isValidSudoku(vector<vector<char>>& board) 
        bool rows[9][9];
        bool columns[9][9];
        bool subBoxes[3][3][9];
        memset(rows, 0, sizeof(rows));
        memset(columns, 0, sizeof(columns));
        memset(subBoxes, 0, sizeof(subBoxes));

        for (int i = 0; i < 9; i++) 
            for (int j = 0; j < 9; j++) 
                char c = board[i][j];
                if (c != '.') 
                    int index = c - '1';

                    if (rows[i][index] || columns[j][index] || subBoxes[i / 3][j / 3][index]) 
                        return false;
                    

                    rows[i][index] = true;
                    columns[j][index] = true;
                    subBoxes[i / 3][j / 3][index] = true;
                
            
        

        return true;
    
;

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c

bool isValidSudoku(char** board, int boardSize, int* boardColSize)
    bool rows[9][9];
    bool columns[9][9];
    bool subBoxes[3][3][9];
    memset(rows, 0, sizeof(rows));
    memset(columns, 0, sizeof(columns));
    memset(subBoxes, 0, sizeof(subBoxes));

    for (int i = 0; i < 9; i++) 
        for (int j = 0; j < 9; j++) 
            char c = board[i][j];
            if (c != '.') 
                int index = c - '1';

                if (rows[i][index] || columns[j][index] || subBoxes[i / 3][j / 3][index]) 
                    return false;
                

                rows[i][index] = true;
                columns[j][index] = true;
                subBoxes[i / 3][j / 3][index] = true;
            
        
    

    return true;

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python

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        rows, columns, sub_boxes = [[False] * 9 for _ in range(9)], [[False] * 9 for _ in range(9)], [
            [[False] * 9 for _ in range(3)] for _ in range(3)]
        for i in range(9):
            for j in range(9):
                c = board[i][j]
                if c != '.':
                    index = ord(c) - ord('1')
                    if rows[i][index] or columns[j][index] or sub_boxes[i // 3][j // 3][index]:
                        return False
                    rows[i][index] = True
                    columns[j][index] = True
                    sub_boxes[i // 3][j // 3][index] = True
        return True

---

java

class Solution 
    public boolean isValidSudoku(char[][] board) 
        boolean[][]   rows     = new boolean[9][9];
        boolean[][]   columns  = new boolean[9][9];
        boolean[][][] subBoxes = new boolean[3][3][9];

        for (int i = 0; i < 9; i++) 
            for (int j = 0; j < 9; j++) 
                char c = board[i][j];
                if (c != '.') 
                    int index = c - '1';

                    if (rows[i][index] || columns[j][index] || subBoxes[i / 3][j / 3][index]) 
                        return false;
                    

                    rows[i][index] = true;
                    columns[j][index] = true;
                    subBoxes[i / 3][j / 3][index] = true;

                
            
        

        return true;
    

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