Deletion is insertion in reverse - find the predecessor at each level, then unlink the target from every level its tower reaches. Today you build the core of delete, splicing a node out cleanly wherever it appears.
Remove a key by unlinking its node from each level of its tower using the update array.
Deletion reuses the exact descent that insertion does. You walk down the levels
building the update array of predecessors, then look at the node just ahead on
level 0. If it is the key you want gone, you unlink it: for every level, if the
predecessor there points straight at the target, redirect that pointer to skip over
it to whatever the target pointed at. A height-3 node like 5 is spliced out of all
three of its levels in one pass, and shorter towers ahead of it are untouched.
The guard update[i].forward[i] == x matters: a predecessor only needs rewiring on
the levels where it actually points at the target. On higher levels above the
target’s tower, update[i] points somewhere else entirely and must be left alone.
Deleting 5 drops it from levels 0, 1, and 2, leaving 1 and 4 as the remaining tall
towers. One thing this does not yet handle is the list level dropping when the top
lane empties out - that is the next lesson.
// Same descent to build update[]. Then, if the successor is the target,// for each level whose predecessor points AT the target, point past it.x = x.forward[0]if x != nil && x.key == key {for i := 0; i < s.level; i++ {if update[i].forward[i] == x { update[i].forward[i] = x.forward[i] }}s.length--}