Longest run of a character in a string

05AB1E, 5 bytes

Code:

SQγOM

Uses the 05AB1E encoding. Try it online!

Explanation:

SQ      # Check for each character if it is equal to the second input
  γ     # Split the list of zeros and ones into groups
   O    # Sum each array in the arrays
    M   # Get the maximum

Mathematica, 35 bytes

Max[Tr/@Split@Boole@Thread[#==#2]]&

Pure function taking a list of characters and another character as input and returning a nonnegative integer. Improved upon my first effort using Adnan's observation (go upvote!) that one should test for equaling the special character before splitting the array.

Thread[#==#2] checks whether each input character in the first argument equals the character given as the second argument. Boole converts the resulting Trues and Falses to 1s and 0s. Split splits the list into runs of consecutive elements; Tr/@ sums each sublist, and Max finds the winner. (Because of how Max works, if the first argument is the empty list, then this function returns -∞. So, you know, don't do that.)

first submission (51 bytes)

Max[Split@#/.a:{c_String..}:>Boole[c==#2]Length@a]&

Split@# splits the input into runs of consecutive characters, such as {{"t"}, {"h"}, {"r"}, {"e", "e"}, {" ", " ", " "}, {"s"}, {"p"}, {"a"}, {"c"}, {"e"}, {"s"}} for the fourth test case. /.a:{c_String..}:> replaces each subexpression a that is a list of a repeated character c by Length@a multiplied by Boole[c==#2], which is 1 if c equals the input character and 0 otherwise. Then Max extracts the answer.

Japt, 20 18 15 bytes

fV+Vî+)ª0)n o l

Try it online!

Saved 5 bytes thanks to obarakon and ETHproductions