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LeetCode題目--有效的數獨(python/Java實現)

題目

判斷一個 9x9 的數獨是否有效。只需要根據以下規則,驗證已經填入的數字是否有效即可。

  1. 數字 1-9 在每一行只能出現一次。
  2. 數字 1-9 在每一列只能出現一次。
  3. 數字 1-9 在每一個以粗實線分隔的 3x3 宮內只能出現一次。

上圖是一個部分填充的有效的數獨。

數獨部分空格內已填入了數字,空白格用 '.' 表示。

示例 1:

輸入:
[
  ["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"]
]
輸出:
true

示例 2:

輸入:
[
  ["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"]
]
輸出:
false 解釋: 除了第一行的第一個數字從 5 改為 8 以外,空格內其他數字均與 示例1 相同。 但由於位於左上角的 3x3 宮內有兩個 8 存在, 因此這個數獨是無效的。

說明:

  • 一個有效的數獨(部分已被填充)不一定是可解的。
  • 只需要根據以上規則,驗證已經填入的數字是否有效即可。
  • 給定數獨序列只包含數字 1-9 和字元 '.' 。
  • 給定數獨永遠是 9x9 形式的。

python語言程式碼實現:

class Solution:
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        for i in range(0,9):
            row = [0] * 9
            col = [0] * 9
            cube = [0] * 9
            for j in range(0,9):
                if board[i][j] !='.':
                    if row[int(board[i][j])-1] == 1:
                        print(1)
                        return False
                    else:
                        row[int(board[i][j])-1] = 1
                if board[j][i] !='.':
                    if col[int (board[j][i])-1] == 1:
                        print(2)
                        return False
                    else:
                        col[int (board[j][i])-1] = 1
                cubeX = int (3*(i//3)+j//3)
                cubeY = int (3 * (i % 3) + j % 3)
                if board[cubeX][cubeY] !='.':
                    if cube[int(board[cubeX][cubeY])-1] == 1:
                        print(cubeX)
                        print(cubeY)
                        print(int(board[cubeX][cubeY])-1)
                        return False
                    else:
                        cube[int (board[cubeX][cubeY])-1] = 1
        return True
        

Java語言程式碼實現:

class Solution {
        public boolean isValidSudoku(char[][] board) {
            //點陣圖法
            //維護三個陣列:行,列、宮
            for(int i=0; i<9; i++){
                int[] row = new int[9];
                int[] col = new int[9];
                int[] cube = new int[9];

                for (int j=0; j<9; j++){
                    if(board[i][j]!='.'){
                        if(row[board[i][j] - '1'] ==1){
                            return false;
                        }else {
                            row[board[i][j] - '1'] =1;
                        }
                    }
                    if(board[j][i] != '.'){
                        if(col[board[j][i] - '1'] == 1){
                            return false;
                        }else{
                            col[board[j][i] - '1'] = 1;
                        }
                    }
                    //每一宮內行列的變換
                    int cubeX = 3 * (i/3) + j/3;
                    int cubeY = 3 * (i%3) + j%3;
                    if(board[cubeX][cubeY]!='.'){
                        if  (cube[board[cubeX][cubeY] - '1'] == 1){
                            return false;
                        }else{
                            cube[board[cubeX][cubeY] - '1'] = 1;
                        }
                    }
                }
            }
            return true;
        }
    }