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LeetCode 面试经典150题 [33/150 有效的数独]
题目描述
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。 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。（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 1：
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个人C++解答
这边建议看官方题解，我写的就是shi，不过对于数独来说够用了，毕竟只有9个数； 官方题解：Leetcode【36.有效的数独】题解 https://leetcode.cn/problems/valid-sudoku/solutions/1001859/you-xiao-de-shu-du-by-leetcode-solution-50m6/
class Solution {public: bool isValidSudoku(vector<vector<char>>& board) { int lin[10]; int col[10]; int reg[10]; for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { if (board[i][j] != '.') { if (lin[board[i][j] - '0'] != 0) return false; else lin[board[i][j] - '0']++; } } fill(lin, lin + 10, 0); } for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { if (board[j][i] != '.') { if (col[board[j][i] - '0'] != 0) return false; else col[board[j][i] - '0']++; } } fill(col, col + 10, 0); }
for (int ii = 0; ii < 7; ii += 3) { for (int jj = 0; jj < 7; jj += 3) { for (int i = ii; i < ii + 3; i++) { for (int j = jj; j < jj + 3; j++) { if (board[i][j] != '.') { if (reg[board[i][j] - '0'] != 0) return false; else reg[board[i][j] - '0']++; } } } fill(reg, reg + 10, 0); } } return true; }}; 分享
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LeetCode 面试经典150题 [33/150 有效的数独]
https://ericonline.cn/posts/leetcode-面试经典150题-33-150-有效的数独/ 部分信息可能已经过时
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