leetcode36.有效的数独

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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：

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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

分析：

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依次按行、列，单元格枚举

代码

  • C++

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class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        bool st[9];
		// 按行枚举
        for (int i = 0;i < 9;i ++) {
            memset(st, false, sizeof st);
            for (int j = 0;j < 9;j ++) {
                if (board[i][j] == '.') continue;
                int t = board[i][j] - '1';
                if (st[t]) return false;
                st[t] = true;
            }
        }
		// 按列枚举
        for (int i = 0;i < 9;i ++) {
            memset(st, false, sizeof st);
            for (int j = 0;j < 9;j ++) {
                if (board[j][i] == '.') continue;
                int t = board[j][i] - '1';
                if (st[t]) return false;
                st[t] = true;
            }
        }
		// 按单元格枚举
        // 通过左上角坐标枚举单元格
        for (int i = 0;i < 9;i +=3) {
            for (int j = 0;j < 9;j += 3) {
                memset(st, false, sizeof st);
                for (int x = 0;x < 3;x ++)
                    for (int y = 0;y < 3;y ++)
                    if (board[i + x][j + y] != '.') {
                        int t = board[i + x][j + y] - '1';
                        if (st[t]) return false;
                        st[t] = true;
                    }
            }
        }

        return true;

    }
};

[原题链接](36. 有效的数独 - 力扣（Leetcode）)

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