leetcode 36

36. Valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

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

                       

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.

判断数独是否有效，即每一行数不能重复，每一列数不能重复，每个9宫格数不能重复。

思路比较简单，但在实现的过程中遇到一些麻烦。刚开始用最简单的比较法来判断数是否重复，结果运行时间超时。

后采用map容器，大大减少了判断重复的时间。

代码如下：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int i,j,k,l,map[10];
        if(board.size()!=9 || board[0].size()!=9)
        {
            return false;
        }
        for(i=0;i<9;i++){
            memset(map,0,sizeof(map));
            for(j=0;j<9;j++){
                if(board[i][j]=='.')
                {
                   continue; 
                }
                if(board[i][j]<'0' || board[i][j]>'9')
                {
                    return false;
                }
                int num=board[i][j]-'0';
                if(map[num])
                {
                    return false;
                }
                map[num]=1;
            }
        }
        for(j=0;j<9;j++){
            memset(map,0,sizeof(map));
            for(i=0;i<9;i++){
                if(board[i][j]=='.')
                {
                    continue;
                }
                int num=board[i][j]-'0';
                if(map[num])
                {
                    return false;
                }
                map[num]=1;
            }
        }
        for(i=0;i<9;i+=3){
            for(j=0;j<9;j+=3){
                memset(map,0,sizeof(map));
                for(k=i;k<i+3;k++){
                    for(l=j;l<j+3;l++){
                        if(board[k][l]=='.')
                        {
                            continue;
                        }
                        int num=board[k][l]-'0';
                        if(map[num])
                        {
                            return false;
                        }
                        map[num]=1;
                    }
                }
            }
        }
        return true;
    }
};