A four-digit snakelike number satisfies $a_1 \lt a_2 \gt a_3 \lt a_4.$ First count the arrangements of any four distinct digits $w \lt x \lt y \lt z$ into this pattern. The largest digit $z$ must sit in position $2$ or $4.$ If $z$ is in position $4,$ the other three form $a_1 \lt a_2 \gt a_3,$ so the largest of them takes position $2$ and the remaining two can go in either order: $2$ ways. If $z$ is in position $2,$ any of the other three digits can be $a_1,$ and then $a_3 \lt a_4$ fixes the rest: $3$ ways. So each set of four digits admits exactly $5$ snakelike orders.

If $0$ is not among the digits, all $5$ orders give valid numbers: $\binom{9}{4} \cdot 5 = 630.$ If $0$ is among them, note $0$ must occupy position $1$ or $3$ (positions $2$ and $4$ must exceed a neighbor), and position $1$ is forbidden. With $0$ in position $3,$ any of the other three digits can be $a_4,$ and $a_1 \lt a_2$ fixes the rest, so $3$ of the $5$ orders survive: $\binom{9}{3} \cdot 3 = 252.$

The total is $630 + 252 = 882.$