LeetCode解题笔记 6 —— 36.有效的数独

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题目

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。

解法

class Solution {
    public boolean isValidSudoku(char[][] board) {
        if(board.length==0){
            return true;
        }
        List<Set<Character>> vertical = new ArrayList<>();//列Set
        List<Set<Character>> horizontal = new ArrayList<>();//行Set
        for(int i = 0; i< board.length; i++){
            vertical.add(new HashSet<>());
            horizontal.add(new HashSet<>());
        }
        List<Set<Character>> list = new ArrayList<>();//九宫格Set
        for(int i = 0; i< board.length; i++){
            if(i%3 == 0){//每三列清除一次
                list.clear();
                for(int k = 0; k < board.length; k+=3){
                    list.add(new HashSet<>());
                }
            }
           
            for(int j = 0; j < board[i].length; j++){
                int t = j/3;
                char c = board[i][j];
                if(c=='.') continue;
                if(vertical.get(j).contains(c)){
                    return false;
                }
                vertical.get(j).add(c);
                if(horizontal.get(i).contains(c)){
                    return false;
                }
                horizontal.get(i).add(c);
                if(list.get(t).contains(c)){
                    return false;
                }
                list.get(t).add(c);
            }
        }
        return true;
    }
}

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