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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.
public class Solution { public boolean isValidSudoku(char[][] board) { //本题思路:先每行每列判断是否有重复的数,此题借助arraylist,注意clear,contains,add等方法,当board中的元素不为‘.’时,才进行判断 //判断完行和列,判断每个3*3的块。注意遍历的方式,四个for循环 int len=board.length; List<Character> row=new ArrayList<Character>(); List<Character> col=new ArrayList<Character>(); for(int i=0;i<len;i++){ row.clear(); col.clear(); for(int j=0;j<len;j++){ char r=board[i][j]; if(r!=‘.‘){ if(row.contains(r)) return false; else{ row.add(r); } } char c=board[j][i]; if(c!=‘.‘){ if(col.contains(c)) return false; else{ col.add(c); } } } } ArrayList<Character> block=new ArrayList<Character>(); for(int k=0;k<len;k=k+3){//k代表块的行,s代表块的列。每个块再进行遍历 for(int s=0;s<len;s=s+3){ block.clear(); for(int i=0;i<3;i++){ for(int j=0;j<3;j++){ char b=board[i+k][j+s]; if(b!=‘.‘){ if(block.contains(b)) return false; else{ block.add(b); } } } } } } return true; } }
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原文地址:http://www.cnblogs.com/qiaomu/p/4634465.html