Valid Sudoku
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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.
public class Solution { public boolean isValidSudoku(char[][] board) { if(board==null) return false; int row = board.length; int col = row; for(int i=0;i<row;i++){ //数组默认初始化为false,用hashset来判断是否已有这个数也行 boolean[] usedLine = new boolean[row]; for(int j=0;j<row;j++){ if(board[i][j]=='.'){ continue; }else if(board[i][j]!='.'&&usedLine[board[i][j]-'1']==false){ usedLine[board[i][j]-'1']=true; }else{ return false; } } } for(int i=0;i<row;i++){ //数组默认初始化为false boolean[] usedLine = new boolean[row]; for(int j=0;j<row;j++){ if(board[j][i]=='.'){ continue; }else if(board[j][i]!='.'&&usedLine[board[j][i]-'1']==false){ usedLine[board[j][i]-'1']=true; }else{ return false; } } } for(int i=0;i<row;i=i+3){ for(int j=0;j<row;j=j+3){ boolean[] usedLine = new boolean[row]; for(int m=0;m<row/3;m++){ for(int n=0;n<row/3;n++){ if(board[i+m][j+n]=='.'){ continue; }else if(board[i+m][j+n]!='.'&&usedLine[board[i+m][j+n]-'1']==false){ usedLine[board[i+m][j+n]-'1']=true; }else{ return false; } } } } } return true; }}2层for循环参考博客:
http://www.tuicool.com/articles/zMn2ei
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