有效的数独

问题描述

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。

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/valid-sudoku

思路

• vector<vector<int>> rows(9, vector<int>(9, 0)); 每行的数字1-9未出现过
• vector<vector<int>> cols(9, vector<int>(9, 0)); 每列的数字1-9未出现过
• vector<vector<int>> cell(9, vector<int>(9, 0)); 每个3x3 宫格内数字1-9未出现过
  在遍历的过程中，i行的数字num（char）第一次出现时，将rows[i]中第num-'1'+1列的值即rowsi置为1，如果在i行第二次出现即rowsi==1，则不是有效的数独，返回false;
  其他两种，同理。

其中非常有意思的一点是，将9*9的宫格转化为9个3*3的宫格时，每个3*3的宫格占cell中的一行，在遍历i、j时，用(i/3)*3+j/3的值即可计算出board[i][j]在第几个宫格。

题解

class Solution {
public:
    bool isValidSudoku(vector<vector<char>> &board) {
        vector<vector<int>> rows(9, vector<int>(9, 0));
        vector<vector<int>> cols(9, vector<int>(9, 0));
        vector<vector<int>> cell(9, vector<int>(9, 0));
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] == '.') continue;
                int e = board[i][j] - '1';
                if (rows[i][e] == 0) rows[i][e] = 1;
                else return false;
                if (cols[j][e] == 0) cols[j][e] = 1;
                else return false;
                if(cell[(i/3)*3+j/3][e]==0) cell[(i/3)*3+j/3][e]=1;
                else return false;
            }
        }
        return true;
    }
};