#Leetcode# 36. Valid Sudoku

https://leetcode.com/problems/valid-sudoku/

 

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1. Each row must contain the digits 1-9 without repetition.
2. Each column must contain the digits 1-9 without repetition.
3. Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.

A partially filled sudoku which is valid.

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

Example 1:

Input:
[
  ["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"]
]
Output: true

Example 2:

Input:
[
  ["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"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being 
    modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Note:

• A Sudoku board (partially filled) could be valid but is not necessarily solvable.
• Only the filled cells need to be validated according to the mentioned rules.
• The given board contain only digits 1-9 and the character '.'.
• The given board size is always 9x9.

代码：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        map<char, int> mp;
        bool flag = true;
        for(int i = 0; i < 9; i ++) {
            mp.clear();
            for(int j = 0; j < 9; j ++) {
                mp[board[i][j]] ++;
                for(int k = 1; k <= 9; k ++) {
                    if(mp['0' + k] > 1) {
                        flag = false;
                        break;
                    }
                }
            }
            mp.clear();
            for(int j = 0; j < 9; j ++) {
                mp[board[j][i]] ++;
                for(int k = 1; k <= 9; k ++) {
                    if(mp['0' + k] > 1) {
                        flag = false;
                        break;
                    }
                }
            }
        }
        
        mp.clear();
        for(int i = 0; i < 9 ; i += 3) {
            for(int j = 0; j < 9; j += 3) {
                mp.clear();
                for(int l = i; l < i + 3; l ++) {
                    for(int r = j; r < j + 3; r ++) {
                        mp[board[l][r]] ++;
                        for(int k = 1; k <= 9; k ++) {
                            if(mp['0' + k] > 1) {
                                flag = false;
                                break;
                            }
                        }
                    }
                }
            }
        }
        return flag;
    }
};

写了无数个 $for$ 循环的时候就知道时间复杂度高的不行了。。。题目写的可清楚但是第一次写的时候还是没看到每个 $3 * 3$ 小方块里面也要互不相同

 $vector$ 啊 一直不怎么会用 emmmm 先去看看 $vector$ 再继续撸题 8 ！

 

题解代码：

class Solution {
public:
    bool isValidSudoku(vector<vector<char> > &board) {
        if (board.empty() || board[0].empty()) return false;
        int m = board.size(), n = board[0].size();
        vector<vector<bool> > rowFlag(m, vector<bool>(n, false)); // row[m][n] 初始化为 false
        vector<vector<bool> > colFlag(m, vector<bool>(n, false));
        vector<vector<bool> > cellFlag(m, vector<bool>(n, false));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] >= '1' && board[i][j] <= '9') {
                    int c = board[i][j] - '1';
                    if (rowFlag[i][c] || colFlag[c][j] || cellFlag[3 * (i / 3) + j / 3][c]) return false;
                    rowFlag[i][c] = true;
                    colFlag[c][j] = true;
                    cellFlag[3 * (i / 3) + j / 3][c] = true;
                }
            }
        }
        return true;
    }
};