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You have a singly-linked list and want to check if it contains a cycle.
A singly-linked list is built with nodes, where each node has:
For example:
A cycle occurs when a node’s next points back to a previous node in the list. The linked list is no longer linear with a beginning and end—instead, it cycles through a loop of nodes.
Write a function containsCycle that takes the first node in a singly-linked list and returns a boolean indicating whether the list contains a cycle.
Careful—a cycle can occur in the middle of a list, or it can simply mean the last node links back to the first node. Does your function work for both?
We can do this in time and space!
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time and space.
The runtime analysis is a little tricky. The worst case is when we do have a cycle, so we don't return until $fastRunner equals $slowRunner. But how long will that take?
First, we notice that when both runners are circling around the cycle $fastRunner can never skip over $slowRunner. Why is this true?
Suppose $fastRunner had just skipped over $slowRunner. $fastRunner would only be 1 node ahead of $slowRunner, since their speeds differ by only 1. So we would have something like this:
What would the step right before this "skipping step" look like? $fastRunner would be 2 nodes back, and $slowRunner would be 1 node back. But wait, that means they would be at the same node! So $fastRunner didn't skip over $slowRunner! (This is a proof by contradiction.)
Since $fastRunner can't skip over $slowRunner, at most $slowRunner will run around the cycle once and $fastRunner will run around twice. This gives us a runtime of .
For space, we store two variables no matter how long the linked list is, which gives us a space cost of .
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