25S-Section-Solution

25S-Section-Solution - CS106X Autumn 2010 Handout 25S...

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CS106X Handout 25S Autumn 2010 October 27 th – 29 th , 2010 Section Solution Solution 1: Removing Duplicates void RemoveDuplicates(node *list) { for (node *cur = list; cur != NULL; cur = cur->next) { if (cur->next != NULL && cur->value == cur->next->value) { // match? node *duplicate = cur->next; // remember cur->next = cur->next->next; // circumvent delete duplicate; // dispose } } } Solution 2: Braided Lists struct node { int value; node *next; }; The recursive approach is easier to code up, but requires more a much more clever algorithm. void Braid(node *list) { Queue<int> numbers; BraidRec(list, numbers); } void BraidRec(node *list, Queue<int>& numbers) { if (list == NULL) return; numbers.enqueue(list->value); BraidRec(list->next, numbers); node *newNode = new node; newNode->value = numbers.dequeue(); newNode->next = list->next; list->next = newNode; } A fully iterative approach is best handled in two passes: void Build(node *list) { node *reverse = NULL; for (node *curr = list; curr != NULL; curr = curr->next) { node *newNode = new node; newNode->value = curr->value; newNode->next = reverse; reverse = newNode; }
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25S-Section-Solution - CS106X Autumn 2010 Handout 25S...

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