LeetCode-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.

自己的解答：

public boolean isValidSudoku(char[][] board) {    for (int i = 0; i < 9; i++) {    Map<Character, Integer> map = getMap();    for (int j = 0; j < 9; j++) {    if (board[i][j] != '.') {    Integer times = map.get(new Character(board[i][j]));    if (times == 0) {    return false;    }    map.put(new Character(board[i][j]), 0);    }    }    }    for (int i = 0; i < 9; i++) {    Map<Character, Integer> map = getMap();    for (int j = 0; j < 9; j++) {    if (board[j][i] != '.') {    Integer times = map.get(new Character(board[j][i]));    if (times == 0) {    return false;    }    map.put(new Character(board[j][i]), 0);    }    }    }    for (int i = 0; i < 9; i += 3) {    for (int j = 0; j < 9; j += 3) {    Map<Character, Integer> map = getMap();    for (int k = i; k < i+3; k++) {    for (int l = j; l < j+3; l++) {    if (board[k][l] != '.') {        Integer times = map.get(new Character(board[k][l]));        if (times == 0) {        return false;        }        map.put(new Character(board[k][l]), 0);        }    }    }    }    }        return true;    }    public Map<Character, Integer> getMap() {    Map<Character, Integer> list = new HashMap<Character, Integer>();    for (int i = 0; i < 10; i++) {    list.put(new Character((char) (i+'0')), 1);    }    return list;    }

代码比较，但是比较容易理解：

在贴几个比较好的优化：

class Solution{public:    bool isValidSudoku(vector<vector<char> > &board)    {        int used1[9][9] = {0}, used2[9][9] = {0}, used3[9][9] = {0};        for(int i = 0; i < board.size(); ++ i)            for(int j = 0; j < board[i].size(); ++ j)                if(board[i][j] != '.')                {                    int num = board[i][j] - '0' - 1, k = i / 3 * 3 + j / 3;                    if(used1[i][num] || used2[j][num] || used3[k][num])                        return false;                    used1[i][num] = used2[j][num] = used3[k][num] = 1;                }        return true;    }};

空间换时间了。下面这个把我的map数据结构换成了set。貌似有提高：

public boolean isValidSudoku(char[][] board) {    for (int i=0; i<9; i++) {        if (!isParticallyValid(board,i,0,i,8)) return false;        if (!isParticallyValid(board,0,i,8,i)) return false;    }    for (int i=0;i<3;i++){        for(int j=0;j<3;j++){            if (!isParticallyValid(board,i*3,j*3,i*3+2,j*3+2)) return false;        }    }    return true;}private boolean isParticallyValid(char[][] board, int x1, int y1,int x2,int y2){    Set singleSet = new HashSet();    for (int i= x1; i<=x2; i++){        for (int j=y1;j<=y2; j++){            if (board[i][j]!='.') if(!singleSet.add(board[i][j])) return false;        }    }    return true;}

但是能不能，只用一个位标识呢？位操作，更加高效，简洁：

bool isValidSudoku(vector<vector<char> > &board) {        int rows = board.size();        int cols = board[0].size();        int mask=0;        int val;        for(int i=0;i<rows;i++){            mask=0;            for(int j=0;j<cols;j++){                if(board[i][j]!='.'){                    int val = (int)(board[i][j]-'0');                    if(mask & (1<<val))                        return false;                    else mask |= (1<<val);                }            }        }        //rows are checked        for(int j=0;j<cols;j++){            mask=0;            for(int i=0;i<rows;i++){                if(board[i][j]!='.'){                    int val = (int)(board[i][j]-'0');                    if(mask & (1<<val))                        return false;                    else mask |= (1<<val);                }            }        }        //cols are checked        for(int i=0;i<rows;i+=3)        {            for(int j=0;j<cols;j+=3)            {                mask=0;                for(int k=i;k<=i+2;k++)                {                    for(int l=j;l<=j+2;l++)                    {                        if(board[k][l]!='.'){                            val = (int)(board[k][l]-'0');                            if(mask&(1<<val))                                return false;                            else mask|=(1<<val);                        }                    }                }            }        }        return true;    }

我喜欢！！！