题目：

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 '.'.![这里写图片描写叙述](http://img.blog.csdn.net/20160409183641502)A partially filled sudoku which is valid.A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

思路：

• 首先这是一道数独题目，关于数独的性质。能够參考以下的链接

• 考虑依据数独的各个条件来逐个解决。须要推断每行，每一列。以及每一个小方格是否是1~9，出现一次。思路是建立一个list加入。推断是否出现反复。出现返回false；相同假设board == null。或者行数和列数不等于9，也返回false

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代码：

public class Solution {    public boolean isValidSudoku(char[][] board) {        ArrayList
      
       
        > rows = new ArrayList
        
         
          >();        ArrayList
          
           
            > cols = new ArrayList
            
             
              >(); ArrayList
              
               
                > boxs = new ArrayList
                
                 
                  >(); if(board == null || board.length != 9|| board[0].length != 9){ return false; } for(int i = 0;i < 9;i++){ rows.add(new ArrayList
                  
                   ()); cols.add(new ArrayList
                   
                    ()); boxs.add(new ArrayList
                    
                     ()); } for(int a = 0;a < board.length;a++){ for(int b = 0;b < board[0].length;b++){ if(board[a][b] == '.'){ continue; } ArrayList
                     
                       row = rows.get(a); if(row.contains(board[a][b])){ return false; }else{ row.add(board[a][b]); } ArrayList
                      
                        col = cols.get(b); if(col.contains(board[a][b])){ return false; }else{ col.add(board[a][b]); } ArrayList
                       
                         box = boxs.get(getNum(a,b)); if(box.contains(board[a][b])){ return false; }else{ box.add(board[a][b]); } } }return true; } public int getNum(int i,int j){ return (i/3)*3+j/3; }}