CSDN公式左对齐，以及遇到的问题及解决方法

CSDN插入公式左对齐方法：

$\begin{array}{r}{h}_{\theta }\left(x\right)={\theta }_0+{\theta }_1{x}_1+{\theta }_2{x}_2\end{array}$
正常情况下，如上面所示已经实现了左对齐，源码为：

$\begin{aligned}
h_{\theta}\left( x \right) =\theta _0+\theta _1x_1+\theta _2x_2
\end{aligned}$

$$\begin{array}{r}{h}_{\theta }\left(x\right)={\theta }_0+{\theta }_1{x}_1+{\theta }_2{x}_2\end{array}$$
正常情况下，如上面所示已经实现了中心对齐，源码为：

$$\begin{aligned}
h_{\theta}\left( x \right) =\theta _0+\theta _1x_1+\theta _2x_2
\end{aligned}$$

区别为左对齐为$ 中心对齐为 $$

但是理想很丰满，现实很骨感

实际使用中遇到的问题：

问题描述：我从Axmath直接复制文本代码，也是LaTeX格式的，但是按照上面操作时，会出现右对齐的现象，如下面所示：
$\begin{array}{r}p\left({\epsilon }^{\left(i\right)}\right)=\frac{1}{\sqrt{2\pi \sigma }}\exp \left(-\frac{{\left({\epsilon }^{\left(i\right)}\right)}^2}{2{\sigma }^2}\right)\\ \\ \because y^{\left(i\right)}={\theta }^T{x}^{\left(i\right)}+{\epsilon }^{\left(i\right)}\\ \\ p\left(y^{\left(i\right)}|x^{\left(i\right)};\theta \right)=\frac{1}{\sqrt{2\pi \sigma }}\exp \left(-\frac{{\left(y^{\left(i\right)}-{\theta }^T{x}^{\left(i\right)}\right)}^2}{2{\sigma }^2}\right)\\ \end{array}$
源码为：

$\begin{aligned}
p\left( \varepsilon ^{\left( i \right)} \right) =\frac{1}{\sqrt{2\pi \sigma}}\exp \left( -\frac{\left( \varepsilon ^{\left( i \right)} \right) ^2}{2\sigma ^2} \right) \\\\
\because y^{\left( i \right)}=\theta ^Tx^{\left( i \right)}+\varepsilon ^{\left( i \right)}\\\\
p\left( y^{\left( i \right)}|x^{\left( i \right)};\theta \right) =\frac{1}{\sqrt{2\pi \sigma}}\exp \left( -\frac{\left( y^{\left( i \right)}-\theta ^Tx^{\left( i \right)} \right) ^2}{2\sigma ^2} \right) \\\\
\end{aligned}$

找遍全网，最后在我的误操作下勉强解决了这个问题，只不过最后会多个字符.：
$$\begin{gathered} p\left({\epsilon }^{\left(i\right)}\right)=\frac{1}{\sqrt{2\pi \sigma }}\exp \left(-\frac{{\left({\epsilon }^{\left(i\right)}\right)}^2}{2{\sigma }^2}\right) \\ \because y^{\left(i\right)}={\theta }^T{x}^{\left(i\right)}+{\epsilon }^{\left(i\right)} \\ p\left(y^{\left(i\right)}|x^{\left(i\right)};\theta \right)=\frac{1}{\sqrt{2\pi \sigma }}\exp \left(-\frac{{\left(y^{\left(i\right)}-{\theta }^T{x}^{\left(i\right)}\right)}^2}{2{\sigma }^2}\right) \\ . \end{gathered}$$
源码为：

$\\
p\left( \varepsilon ^{\left( i \right)} \right) =\frac{1}{\sqrt{2\pi \sigma}}\exp \left( -\frac{\left( \varepsilon ^{\left( i \right)} \right) ^2}{2\sigma ^2} \right) \\\\
\because y^{\left( i \right)}=\theta ^Tx^{\left( i \right)}+\varepsilon ^{\left( i \right)}\\\\
p\left( y^{\left( i \right)}|x^{\left( i \right)};\theta \right) =\frac{1}{\sqrt{2\pi \sigma}}\exp \left( -\frac{\left( y^{\left( i \right)}-\theta ^Tx^{\left( i \right)} \right) ^2}{2\sigma ^2} \right) \\\\
\\.$

最后为了公式看起来更加清晰，在行与行之间加一行空白行，效果如下：
$$\begin{gathered} p\left({\epsilon }^{\left(i\right)}\right)=\frac{1}{\sqrt{2\pi \sigma }}\exp \left(-\frac{{\left({\epsilon }^{\left(i\right)}\right)}^2}{2{\sigma }^2}\right) \\ \because y^{\left(i\right)}={\theta }^T{x}^{\left(i\right)}+{\epsilon }^{\left(i\right)} \\ p\left(y^{\left(i\right)}|x^{\left(i\right)};\theta \right)=\frac{1}{\sqrt{2\pi \sigma }}\exp \left(-\frac{{\left(y^{\left(i\right)}-{\theta }^T{x}^{\left(i\right)}\right)}^2}{2{\sigma }^2}\right) \\ . \end{gathered}$$
源码为：

$\\
p\left( \varepsilon ^{\left( i \right)} \right) =\frac{1}{\sqrt{2\pi \sigma}}\exp \left( -\frac{\left( \varepsilon ^{\left( i \right)} \right) ^2}{2\sigma ^2} \right) 
\\\\
~\\
\because y^{\left( i \right)}=\theta ^Tx^{\left( i \right)}+\varepsilon ^{\left( i \right)}\\\\
~\\
p\left( y^{\left( i \right)}|x^{\left( i \right)};\theta \right) =\frac{1}{\sqrt{2\pi \sigma}}\exp \left( -\frac{\left( y^{\left( i \right)}-\theta ^Tx^{\left( i \right)} \right) ^2}{2\sigma ^2} \right) \\\\
\\.$

~\为显示空白行