Spoiler:
The sequence that needs, ON AVERAGE, less coin tosses is THH.
I'll explain you why, trying to be as simple as I can be: each toss is an independent event, which means that each time I toss the coin I have 50% chances to get Head, and 50% chances to get Tail.
Following this logic, each sequence (THH and THT) should have the same chances to happen... BUT! There's one thing to consider: we're looking for specific sequences!
Let's assume that we are now trying to get the THT sequence, and we already got one Tail and one Head (TH). Now we toss the coin, and we have 50% chances to get Tail (and complete the THT), and 50% chances to get Head, so we have to continue tossing: in case we get Head, so we DON'T finish the sequence, we'll need AT LEAST three more tosses, because the latest two results we got would be Head-Head, so in the most lucky condition we would have to get another Tail followed by another Head and another Tail.
Let's assume now that we are trying to get the THH sequence, and once again we already got one Tail and one Head (TH). Once again, we have 50% chances to get Head (thus completing the sequence), and50% chances to get Tail (thus FAILING the sequence). But Wait! If we fail, our last toss resulted in a Tail (T), so in the most lucky condition we only would need TWO more tosses (two Heads) to complete the sequence, instead of three.
This small, unique case makes the combination THH "quicker" than the combination THT on average