【LeetCode】36、有效的数独

36、有效的数独

题目：

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。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 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。

示例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

示例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 ‘.’

解题思路：

乍一看此题，完全没有思路，默默的打开了讨论区，看着大家的天才解法，不由的陷入了一阵深思…

先说一下暴力解法：

1. 先对整个board进行一次遍历，查找是否在每一行以及每一列有相同的元素
2. 再对每个box进行遍历，使用Set来查找每个box中是否有相同的元素
3. 如果不能添加说明有重复元素

参考代码：

  public boolean isValidSudoku(char[][] board) {
    
        for (int i = 0; i < 9; i++){
    
            for (int j = 0; j < 9; j++){
    
                if (board[i][j]!= '.'){
    
                    //判断这一行中是否有相同元素
                    for (int k = j+1; k < 9; k ++){
    
                        if (board[i][j] == board[i][k] ){
    
                            return false;
                        }
                    }

                    //判断这一列中是否有相同元素
                    for (int k = i+1; k<9; k++){
    
                        if (board[i][j] == board[k][j]){
    
                            return false;
                        }
                    }
                }
            }
        }

        //判断每个一box中是否有相同元素
        for (int i = 0; i < 9; i=i+3){
    
            for (int j = 0; j < 9; j = j+3){
    
                Set<Character> set = new HashSet<>();
                for (int k = 0; k <3; k++){
    
                    for (int l = 0; l < 3; l++){
    
                        if (board[i + k][j + l]!= '.' && !set.add(board[i + k][j + l])){
    
                            return false;
                        }
                    }
                }
            }
        }

        return true;
    }
}