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 ‘.‘.
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.
思路：
先验证每一行，再验证每一列，再验证每一个小正方形。
package sudoku; public class ValidSudoku { public boolean isValidSudoku(char[][] board) { int n = 9; for (int i = 0; i < n; ++i) { int[] arr = new int[10]; for (int j = 0; j < n; ++j) { if (board[i][j] != ‘.‘) arr[board[i][j] - ‘0‘]++; } if (!validate(arr)) return false; } for (int i = 0; i < n; ++i) { int[] arr = new int[10]; for (int j = 0; j < n; ++j) { if (board[j][i] != ‘.‘) arr[board[j][i] - ‘0‘]++; } if (!validate(arr)) return false; } for (int i = 0; i < n; i += 3) { for (int j = 0; j < n; j += 3) { int[] arr = new int[10]; for (int x = i; x < i + 3; ++x) { for (int y = j; y < j + 3; ++y) { if (board[x][y] != ‘.‘) arr[board[x][y] - ‘0‘]++; } } if (!validate(arr)) return false; } } return true; } private boolean validate(int[] arr) { for (int i = 1; i < 10; ++i) { if (arr[i] > 1) return false; } return true; } public static void main(String[] args) { // TODO Auto-generated method stub } }
原文：http://www.cnblogs.com/shuaiwhu/p/5068732.html