有效的數(shù)獨

判斷一個 9x9 的數(shù)獨是否有效。只需要根據(jù)以下規(guī)則，驗證已經(jīng)填入的數(shù)字是否有效即可。

數(shù)字 1-9 在每一行只能出現(xiàn)一次。
數(shù)字 1-9 在每一列只能出現(xiàn)一次。
數(shù)字 1-9 在每一個以粗實線分隔的 3x3 宮內只能出現(xiàn)一次。

上圖是一個部分填充的有效的數(shù)獨。

數(shù)獨部分空格內已填入了數(shù)字，空白格用 ‘.’ 表示。

示例 1:

輸入: [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"] ] 輸出: true

示例 2:

輸入: [["8","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"] ] 輸出: false 解釋: 除了第一行的第一個數(shù)字從 5 改為 8 以外，空格內其他數(shù)字均與 示例1 相同。但由于位于左上角的 3x3 宮內有兩個 8 存在, 因此這個數(shù)獨是無效的。

說明:

一個有效的數(shù)獨（部分已被填充）不一定是可解的。
只需要根據(jù)以上規(guī)則，驗證已經(jīng)填入的數(shù)字是否有效即可。
給定數(shù)獨序列只包含數(shù)字 1-9 和字符 ‘.’ 。
給定數(shù)獨永遠是 9x9 形式的。

思路+代碼+注釋：

public static boolean isValidSudoku(char[][] board) {/*思路：先看每一行是否有重復的數(shù)字，每一行就是二維數(shù)組中的元素再看每一列是否有重復的數(shù)字，所有二維數(shù)組元素的同一個位置字符針對3*3的格子(橫著數(shù))第一個格子是外層0 1 2、內層0 1 2第二個格子是外層0 1 2、內層3 4 5第三個格子是外層0 1 2、內層6 7 8第二層三個格子外層都是3 4 5，內層分別是0 1 2，3 4 5，6 7 8*///定義一個大小為9的數(shù)組，數(shù)組中每個元素代表對應數(shù)字出現(xiàn)的次數(shù)int[] numCount=new int[9];//驗證每一行的for (int i = 0; i < board.length; i++) {for (char c:board[i]) {switch (c){case '1':numCount[0]++;break;case '2':numCount[1]++;break;case '3':numCount[2]++;break;case '4':numCount[3]++;break;case '5':numCount[4]++;break;case '6':numCount[5]++;break;case '7':numCount[6]++;break;case '8':numCount[7]++;break;case '9':numCount[8]++;break;}}for (int j = 0; j < numCount.length; j++) {//如果大于1證明出現(xiàn)多次，直接返回falseif (numCount[j]>1){return false;}//如果不大于1，那么歸零，為下次做準備numCount[j]=0;}}//驗證每一列的,共9列for (int i = 0; i < numCount.length; i++) {//每一列for (int j = 0; j < board.length; j++) {char c=board[j][i];switch (c){case '1':numCount[0]++;break;case '2':numCount[1]++;break;case '3':numCount[2]++;break;case '4':numCount[3]++;break;case '5':numCount[4]++;break;case '6':numCount[5]++;break;case '7':numCount[6]++;break;case '8':numCount[7]++;break;case '9':numCount[8]++;break;}}for (int j = 0; j < numCount.length; j++) {//如果大于1證明出現(xiàn)多次，直接返回falseif (numCount[j]>1){return false;}//如果不大于1，那么歸零，為下次做準備numCount[j]=0;}}//驗證三層格子//驗證第一層3個格子int j=0;for (int k = 0; k < 3; k++) {int endJ=j+3;//驗證每個格子for (int i = 0; i < 3; i++) {j=endJ-3;for (; j < endJ; j++) {char c=board[i][j];switch (c){case '1':numCount[0]++;break;case '2':numCount[1]++;break;case '3':numCount[2]++;break;case '4':numCount[3]++;break;case '5':numCount[4]++;break;case '6':numCount[5]++;break;case '7':numCount[6]++;break;case '8':numCount[7]++;break;case '9':numCount[8]++;break;}}}for (int m = 0; m < numCount.length; m++) {//如果大于1證明出現(xiàn)多次，直接返回falseif (numCount[m]>1){return false;}//如果不大于1，那么歸零，為下次做準備numCount[m]=0;}}//驗證第二層3個格子j=0;for (int k = 0; k < 3; k++) {int endJ=j+3;//驗證每個格子for (int i = 3; i < 6; i++) {j=endJ-3;for (; j < endJ; j++) {char c=board[i][j];switch (c){case '1':numCount[0]++;break;case '2':numCount[1]++;break;case '3':numCount[2]++;break;case '4':numCount[3]++;break;case '5':numCount[4]++;break;case '6':numCount[5]++;break;case '7':numCount[6]++;break;case '8':numCount[7]++;break;case '9':numCount[8]++;break;}}}for (int m = 0; m < numCount.length; m++) {//如果大于1證明出現(xiàn)多次，直接返回falseif (numCount[m]>1){return false;}//如果不大于1，那么歸零，為下次做準備numCount[m]=0;}}//驗證第三層3個格子j=0;for (int k = 0; k < 3; k++) {int endJ=j+3;//驗證每個格子for (int i = 6; i < 9; i++) {j=endJ-3;for (; j < endJ; j++) {char c=board[i][j];switch (c){case '1':numCount[0]++;break;case '2':numCount[1]++;break;case '3':numCount[2]++;break;case '4':numCount[3]++;break;case '5':numCount[4]++;break;case '6':numCount[5]++;break;case '7':numCount[6]++;break;case '8':numCount[7]++;break;case '9':numCount[8]++;break;}}}for (int m = 0; m < numCount.length; m++) {//如果大于1證明出現(xiàn)多次，直接返回falseif (numCount[m]>1){return false;}//如果不大于1，那么歸零，為下次做準備numCount[m]=0;}}return true;}

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