leetcode 36. Valid Sudoku

最新推荐文章于 2022-07-31 21:20:38 发布

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最新推荐文章于 2022-07-31 21:20:38 发布

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本文介绍了一种使用C++实现的数独有效性验证算法。该算法通过检查每一行、每一列以及每一个九宫格内的数字是否重复来确定数独是否有效。文章提供了详细的代码实现，并解释了如何检查数独是否符合基本规则。

解题思路：
直接判断，9行，9列，9个子面板是否符合要求

原题目：

Determine if a Sudoku is valid, according to: [Sudoku Puzzles - The Rules.](http://sudoku.com.au/TheRules.aspx)

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

这里写图片描述

A partially filled sudoku which is valid.

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

AC解，C++代码，菜鸟一个，请大家多多指正

class Solution {
public:
    bool isValidNums(vector<char> nums) {
        map<char, int> char_count;
        for (vector<char>::size_type ix = 0; ix != nums.size(); ++ix) {
            if (nums[ix] != '.') {
                char_count[nums[ix]]++;
                if (char_count[nums[ix]] > 1) {
                    return false;
                }
            }
        }
        return true;
    }

    bool isValidSudoku(vector<vector<char>>& board) {
        for (int row = 0; row < 9; row++) {
            if (!isValidNums(board[row])) {
                return false;
            }
            for (int col = 0; col < 9; col++) {
                if (row % 3 == 0 && col % 3 == 0) {
                    vector<char> tmp;
                    tmp.push_back(board[row][col]);
                    tmp.push_back(board[row][col + 1]);
                    tmp.push_back(board[row][col + 2]);
                    tmp.push_back(board[row + 1][col]);
                    tmp.push_back(board[row + 1][col + 1]);
                    tmp.push_back(board[row + 1][col + 2]);
                    tmp.push_back(board[row + 2][col]);
                    tmp.push_back(board[row + 2][col + 1]);
                    tmp.push_back(board[row + 2][col + 2]);
                    if (!isValidNums(tmp)) {
                        return false;
                    }
                }
            }
        }
        for (int col = 0; col < 9; col++) {
            vector<char> tmp;
            for (int row = 0; row < 9; row++) {
                tmp.push_back(board[row][col]);
            }
            if (!isValidNums(tmp)) {
                return false;
            }
        }
        return true;
    }
};