LeetCode 面试经典150题 [33/150 有效的数独]


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GuoYulong 2024-06-25 79

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/

GYL
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;
    }
};

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