作者:力扣 (LeetCode) 链接:https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2f9gg/ 来源:力扣(LeetCode)
题目描述:
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。
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。
示例 1:
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。答案v1.0
class Solution { public: bool isValidSudoku(vector<vector<char>> &board) { //方法1:暴力遍历 for (int i = 0; i < board.size(); i++) { for (int j = 0; j < board[i].size(); j++) { if (board[i][j] != '.') { for (int row = 0; row < board[i].size(); row++) //横行 { if (board[i][j] == board[i][row] && row != j) { return false; } } for (int col = 0; col < board.size(); col++) //竖列 { if (board[i][j] == board[col][j] && col != i) { return false; } } // 3*3小方块 for (int row3 = int(j / 3) * 3; row3 < int(j / 3) * 3 + 3; row3++) { for (int col3 = int(i / 3) * 3; col3 < int(i / 3) * 3 + 3; col3++) { if (board[i][j] == board[col3][row3] && col3 != i && row3 != j) { return false; } } } } } } return true; } };