[leetcode]Valid Sudoku

问题描述：

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.

代码：

public class Valid_Sudoku {  //javapublic boolean isValidSudoku(char[][] board) {        int size = 9;int [] member = new int[size];   //record if is occur; //valid rowfor(int i = 0; i < 9; i++){for(int k = 0; k < 9; k++)member[k] = 0;for(int j = 0; j < 9; j++){if(board[i][j] == '.')continue;int pos = board[i][j]-'0';if(member[pos-1] == 1)return false;else member[pos-1] = 1;}}//valid colfor(int i = 0; i < 9; i++){for(int k = 0; k < 9; k++)member[k] = 0;for(int j = 0; j < 9; j++){if(board[j][i] == '.')continue;int pos = board[j][i]-'0';if(member[pos-1] == 1)return false;else member[pos-1] = 1;}}//valid cubefor(int ibegin = 0; ibegin < 9; ibegin = ibegin+3){for(int jbegin = 0; jbegin < 9; jbegin = jbegin+3){for(int k = 0; k < 9; k++)member[k] = 0;for(int i = ibegin; i < ibegin+3; i++){for(int j = jbegin; j < jbegin+3; j++){if(board[i][j] == '.')continue;int pos = board[i][j]-'0';if(member[pos-1] == 1)return false;else member[pos-1] = 1;}}}}return true;    }public static void main(String [] args){String[] boardStr = {"......5..", ".........", ".........", "93..2.4..", "..7...3..", ".........", "...34....", ".....3...", ".....52.."};char [][] board = new char [9][9];for(int i =0; i< boardStr.length; i++){for(int j = 0; j<boardStr[i].length(); j++){board[i][j] = boardStr[i].charAt(j);}}Valid_Sudoku vs = new Valid_Sudoku();System.out.println(vs.isValidSudoku(board));}}