這題的解法是brute force啥酱，復雜度是O(81 + 81)挡爵。姆泻。。判斷每一行每一列的代碼我的想法跟code ganker的一致：用一個數(shù)組表示map硼补；難點是判斷每個unit是否valid驮肉，這種數(shù)組下標確實很難思考。括勺。需要在紙上模擬一下才能看懂缆八。

    public boolean isValidSudoku(char[][] board) {
        if (board.length != 9 || board[0].length != 9) return false;
        int[] temp = new int[9];
        //判斷每一行
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] != '.') {
                    if (temp[board[i][j] - '1'] != 0) {
                        return false;
                    }
                    temp[board[i][j] - '1'] = 1;
                }
            }
            temp = new int[9];
        }

        //判斷每一列
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[j][i] != '.') {
                    if (temp[board[j][i] - '1'] != 0) {
                        return false;
                    }
                    temp[board[j][i] - '1'] = 1;
                }
            }
            temp = new int[9];
        }

        //判斷9個小units
        for (int block = 0; block < 9; block++) {
            for (int i = block / 3 * 3; i < block / 3 * 3 + 3; i++) {
                for (int j = block % 3 * 3; j < block % 3 * 3 + 3; j++) {
                    if (board[i][j] != '.') {
                        if (temp[board[i][j] - '1'] != 0) {
                            return false;
                        }
                        temp[board[i][j] - '1'] = 1;
                    }
                }
            }
            temp = new int[9];
        }
        return true;

    }

然后看了solutions里的解法，有個人用hashset疾捍，然后用了i j的reuse，很簡潔栏妖，復雜度也稍低一些：

public boolean isValidSudoku(char[][] board) {
    for(int i = 0; i<9; i++){
        HashSet<Character> rows = new HashSet<Character>();
        HashSet<Character> columns = new HashSet<Character>();
        HashSet<Character> cube = new HashSet<Character>();
        for (int j = 0; j < 9;j++){
            if(board[i][j]!='.' && !rows.add(board[i][j]))
                return false;
            if(board[j][i]!='.' && !columns.add(board[j][i]))
                return false;
            int RowIndex = 3*(i/3);
            int ColIndex = 3*(i%3);
            if(board[RowIndex + j/3][ColIndex + j%3]!='.' && !cube.add(board[RowIndex + j/3][ColIndex + j%3]))
                return false;
        }
    }
    return true;
}