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.
Solution:
There are just 3 rules to Sudoku. |
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Each row must have the numbers 1-9 occuring just once.
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Each column must have the numbers 1-9 occuring just once.
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And the numbers 1-9 must occur just once in each of the 9 sub-boxes of the grid.
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一个valid的sudoku只有3个条件：横、竖、小方块内部 满足条件即可。
我老想成还需要对角线上的数也满足条件，这样想是错的！！！
1 public class Solution { 2 public boolean isValidSudoku(char[][] board) { 3 for(int i=0;i<board.length;++i){ 4 for(int j=0;j<board[0].length;++j){ 5 if(board[i][j]==‘.‘){ 6 continue; 7 }else{ 8 char temp=board[i][j]; 9 board[i][j]=‘C‘; //通过将board[i][j]的值改为board内部的其他位置的数 10 boolean b=isValid(board,i,j,temp); //不可能取到的值来方便后边儿的验证。 11 board[i][j]=temp; 12 if(b==false) 13 return false; 14 } 15 } 16 } 17 return true; 18 } 19 20 private boolean isValid(char[][] board, int x, int y, char temp) { 21 // TODO Auto-generated method stub 22 for(int i=0;i<board.length;++i){ //竖着的 23 if(board[i][y]==temp) 24 return false; 25 } 26 for(int i=0;i<board[0].length;++i){ //横着的 27 if(board[x][i]==temp){ 28 return false; 29 } 30 } 31 for(int i=(x/3)*3;i<(x/3+1)*3;++i){ //小方块内部的 32 for(int j=(y/3)*3;j<(y/3+1)*3;++j){ //判断好属于哪块小方格 33 if(board[i][j]==temp) 34 return false; 35 } 36 } 37 return true; 38 } 39 }
原文：http://www.cnblogs.com/Phoebe815/p/4103688.html