LeetCode-36.有效的数独(考察点:HashSet)

判断一个 9x9 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。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字,空白格用 '.' 表示。

示例 1:

输入:
[
  ["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"]
]
输出: true

示例 2:

输入:
[
  ["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"]
]
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。

说明:

  • 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。
  • 给定数独序列只包含数字 1-9 和字符 '.' 。
  • 给定数独永远是 9x9 形式的。

解题思路:HashSet,不可保存重复元素,用3个HashSet,分别保存第i行、第i列和第i个3x3的九宫格中的元素,每处理一个元素,若不为空,将正在处理的当前元素,添加到所属的行、列以及3x3的九宫格中,若添加失败,表明所属的行、列或者3x3九宫格中有重复元素,返回false;若全部扫描完,返回true。

代码如下:

class Solution {
    public boolean isValidSudoku(char[][] board) {
        //最外层循环,每次循环并非只是处理第i行,而是处理第i行、第i列以及第i个3x3的九宫格
        for(int i = 0; i < 9; i++){
            HashSet<Character> line = new HashSet<>();
            HashSet<Character> col = new HashSet<>();
            HashSet<Character> cube = new HashSet<>();
            for(int j = 0; j < 9; j++){
                if('.' != board[i][j] && !line.add(board[i][j]))
                    return false;
                if('.' != board[j][i] && !col.add(board[j][i]))
                    return false;
                int m = i/3*3+j/3;
                int n = i%3*3+j%3;
                if('.' != board[m][n] && !cube.add(board[m][n]))
                    return false;
            }
        }
        return true;
    }
}

猜你喜欢

转载自blog.csdn.net/weixin_38823568/article/details/81202020