Valid Sudoku，validsudoku

Valid Sudoku，validsudoku

本文是在學習中的總結，歡迎轉載但請註明出處：http://blog.csdn.net/pistolove/article/details/42528601

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：

（1）題意為判斷一個數獨是否有效。

（2）題意不是讓我們求解出整個數獨（當然，如果真的求解還是很複雜的），而是判斷數獨中已有資料是否有效。

（3）本文的思路還是暴力破解，因為沒有想到其它好的方法。對數獨對應的9*9的二維數組進行遍曆，對於任意一個不為'.'的數字，都需要對其所在的行、列、以及3*3的塊地區進行判斷，如果有重複出現的情況，那麼數獨就是無效的，具體見下方代碼。

（4）希望本文對你有所協助。

演算法代碼實現如下：

public boolean isValidSudoku(char[][] board) {for (int i = 0; i < board.length; i++) {for (int j = 0; j < board[i].length; j++) {char curr = board[i][j];if(curr=='.'){continue;}//行for (int k = j+1; k < board.length; k++) {if(curr==board[i][k]){return false;}}//列for (int k = i+1; k < board.length; k++) {if(curr==board[k][j]){return false;}}//3*3 方塊if(i>=0&&i<3&&j>=0&&j<3){int count=0;for (int k1 = 0; k1 < 3; k1++) {for (int k2 = 0; k2 < 3; k2++) {if(board[k1][k2]==curr){count++;}if(count>1) return false;}}}if(i>=0&&i<3&&j>=3&&j<6){int count=0;for (int k1 = 0; k1 < 3; k1++) {for (int k2 = 3; k2 < 6; k2++) {if(board[k1][k2]==curr){count++;}if(count>1) return false;}}}if(i>=0&&i<3&&j>=6&&j<9){int count=0;for (int k1 = 0; k1 < 3; k1++) {for (int k2 = 6; k2 < 9; k2++) {if(board[k1][k2]==curr){count++;}if(count>1) return false;}}}if(i>=3&&i<6&&j>=0&&j<3){int count=0;for (int k1 = 3; k1 < 6; k1++) {for (int k2 = 0; k2 < 3; k2++) {if(board[k1][k2]==curr){count++;}if(count>1) return false;}}}if(i>=3&&i<6&&j>=3&&j<6){int count=0;for (int k1 = 3; k1 < 6; k1++) {for (int k2 = 3; k2 < 6; k2++) {if(board[k1][k2]==curr){count++;}if(count>1) return false;}}}if(i>=3&&i<6&&j>=6&&j<9){int count=0;for (int k1 = 3; k1 < 6; k1++) {for (int k2 = 6; k2 < 9; k2++) {if(board[k1][k2]==curr){count++;}if(count>1) return false;}}}if(i>=6&&i<9&&j>=0&&j<3){int count=0;for (int k1 = 6; k1 < 9; k1++) {for (int k2 = 0; k2 < 3; k2++) {if(board[k1][k2]==curr){count++;}if(count>1) return false;}}}if(i>=6&&i<9&&j>=3&&j<6){int count=0;for (int k1 = 6; k1 < 9; k1++) {for (int k2 = 3; k2 < 6; k2++) {if(board[k1][k2]==curr){count++;}if(count>1) return false;}}}if(i>=6&&i<9&&j>=6&&j<9){int count=0;for (int k1 = 6; k1 < 9; k1++) {for (int k2 = 6; k2 < 9; k2++) {if(board[k1][k2]==curr){count++;}if(count>1) return false;}}}}}return true;}