Last digit: Sum of Alternate Sign Squares

$2018^{2}-2017^{2}+2016^{2}-2015^{2}+...+2^{2}-1^{2}$

Group the relevant terms:

$=(2018^{2}-2017^{2})+(2016^{2}-2015^{2})+...+(2^{2}-1^{2})$

From Difference Of Squares :

$=(2018-2017)(2018+2017)+(2016-2015)(2016+2015)+...+(2-1)(2+1)$

Notice the multiplication terms with minus sign all equal to 1:

$=(2018+2017)+(2016+2015)+...+(2+1)$

$=2018+2017+2016+2015+...+2+1$

$=\frac{(2018)\times(2018+1)}{2}$

$=\frac{4074342}{2}=2037171$

Therefore the last digit of the sum is $\boxed{1}$