Question:
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.
Anslysis:
问题描述：判断一个数独游戏是否合法。数独当前可以是部分填充，未填充部分用‘.‘代替。有效地数独并不是指该游戏是否有解，而仅仅判断当前填充是否有效。
数独有效是指：每一行，每一列，以及每个小得九宫格的当前填充是否有重复数字。
思路：看到题目，分析完成后的直接思路是，分别判断数独的每一行，每一列，每九宫格是否含有相同的数字，如果含有就不是有效地数独。
Answer：
public class Solution { public boolean isValidSudoku(char[][] board) { //横向判断 for(int i=0; i<board.length; i++) { HashSet<Character> v = new HashSet<Character>(); for(int j=0; j<board[i].length; j++) { if(board[i][j] != ‘.‘) { if(!v.contains(board[i][j])) v.add(board[i][j]); else return false; } } } //纵向判断 for(int i=0; i<board[0].length; i++) { HashSet<Character> v = new HashSet<Character>(); for(int j=0; j<board.length; j++) { if(board[j][i] != ‘.‘) { if(!v.contains(board[j][i])) v.add(board[j][i]); else return false; } } } //九宫格判断 int i = 0, j = 0; while(i < board.length) { j = 0; while(j < board[0].length) { HashSet<Character> v = new HashSet<Character>(); for(int t=0; t<3; t++) { for(int k=0; k<3; k++) { if(board[i+t][j+k] != ‘.‘) { if(!v.contains(board[i+t][j+k])) v.add(board[i+t][j+k]); else return false; } } } j = j + 3; } i = i + 3; } return true; } }
原文：http://www.cnblogs.com/little-YTMM/p/4855924.html