leetcode 36. Valid Sudoku 37. Sudoku Solver

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.

判断一个数独盘的已经填上的部分是不是有效的数独盘。不需要判断没填上的部分。也就是看有没有重复的。

class Solution {public:    bool isValidSudoku(vector<vector<char>>& board)     {        map<char,int> m2;        //判断每一横排        for (int i = 0; i < 9; i++)        {               for (int j = 0; j < 9; j++)            {                 if (board[i][j] == '.')                     continue;                 if (m2.find(board[i][j]) == m2.end())                     m2[board[i][j]]++;                 else                     return false;            }            m2.clear();         }                  //判断每一竖排        for (int i = 0; i < 9; i++)        {            for (int j = 0; j < 9; j++)            {                if(board[j][i]=='.')                    continue;                if(m2.find(board[j][i])==m2.end())                    m2[board[j][i]]++;                else                    return false;            }            m2.clear();        }        //判断每一个九宫格        for(int k = 0; k < 3; k++)  //k控制 i的范围        {               for(int h = 0; h < 3; h++)    //h控制 j的范围            {                for(int i = k * 3; i < k * 3 + 3; i++)                {                    for(int j = h * 3; j < h * 3 + 3; j++)                    {                        if(board[i][j]=='.')                            continue;                        if(m2.find(board[i][j]) == m2.end())                            m2[board[i][j]]++;                        else                            return false;                    }                }                m2.clear();            }        }          return true;       }};

37. Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

...and its solution numbers marked in red.

填数独
用递归的思想，先判断每一个空位可以填哪些，然后填进去再递归。

class Solution {public:    int shudu(vector<vector<char>>&board,vector<vector<int>> &mm,int pos,int flag)    {        if (pos == mm.size())            return 1;                int a[10] = {0};        int x = mm[pos][0]; //需要填的位置        int y = mm[pos][1];        for(int i = 0; i < 9; i++)        {            if (board[x][i] != '.')//横                a[(int(board[x][i])-48)] = 1;            if (board[i][y] != '.')//竖                a[(int(board[i][y])-48)] = 1;        }        //3*3        for (int k = (x/3)*3; k < (x/3)*3+3; k++)        {            for(int h = (y/3)*3; h<(y/3)*3+3; h++)            {                if(board[k][h]!='.')                    a[(int(board[k][h])-48)] = 1;            }        }        //把可以加的加进去        for(int i = 1; i <= 9; i++)        {            if(a[i] == 0)            {                board[x][y] = char(i+48);                flag = shudu(board,mm,pos+1,flag);                if(flag == 0)   // 填入之后递归失败                    board[x][y] = '.';                else            // 填入之后递归成功                    break;            }        }        return flag;    }        void solveSudoku(vector<vector<char>>& board)    {        //将点('.')的位置存起来        vector<vector<int>> mm;        vector<int> m1;        for (int i = 0; i < 9; i++)        {            for (int j = 0; j < 9; j++)            {                if (board[i][j] == '.')                {                    m1.push_back(i);                    m1.push_back(j);                    mm.push_back(m1);                    m1.clear();                }            }        }        //开始遍历        int k = shudu(board, mm, 0, 0);    }};