12-alt-impl-subtype.student - Last time Extended example of...

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Last time: Extended example of an ADT: the intset Today: * Implementation flexibility Given this implementation, how many elements of the array must indexOf examine in the *worst case* if there are 10 elements? If there are 90 elements? We say the time for indexOf grows *linearly* with the size of the set. If there are N elements in the set, we have to examine all N of them in the worst case. If we have large sets, and are performing lots of queries, this might be too expensive. Luckily for us, we can replace this implementation with a different one that can be more efficient. The only change we need to make is to the representation---the the abstraction can stay precisely the same. We'll still use an array (of 100 elements) to store the elements of the set, and the will still occupy the first numElts slots, with no duplicates. However, this time, rather than unsorted, we'll keep the elements in sorted order. The constructor and size methods don't need to change at all---all they use is the numElts field. Likewise, query doesn't need to change either---if the index exists in the array's legal bounds, then it's there. However, the others all do need to change. We'll start with the easiest one: remove. Recall its old definition: void IntSet::remove(int v) { int victim = indexOf(v); if (victim != NOTFOUND) { numElts--; elts[victim] = elts[numElts]; } } This moves the last (and largest) element from the end to the middle, which breaks the new invariant. So, instead of dong a swap, we have to "squish" the array together to cover up the hole. We'll do that by moving the element next to the hole to the left, and this leaves a new hole. We'll keep moving elements until the hole is
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"off the end" of the elements. We'll re-use the variable victim as a loop variable, its invariant is that it always points at the hole in the array. void IntSet::remove(int v)
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12-alt-impl-subtype.student - Last time Extended example of...

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