1. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{2}=3$, and $a_{n+2}=\left|a_{n+1}-a_{n}\right|\left(n \in \mathbf{Z}_{+}\right)$.
Then $a_{2014}=$ $\qquad$
-1.1 .
It is easy to see that $a_{1}=1, a_{2}=3, a_{3}=2, a_{4}=1, a_{5}=1$,
$$
a_{6}=0, a_{7}=1, a_{8}=1, a_{9}=0, \cdots \cdots
$$

Therefore, starting from the fourth term, every three consecutive terms periodically take the values $1, 1, 0$.
Thus, $a_{2014}=1$.