ShuffleMerge---microsoft's interview question
Given two lists, merge their nodes together to make one list, taking nodes alternately
between the two lists. So ShuffleMerge() with {1, 2, 3} and {7, 13, 1} should yield {1, 7,
2, 13, 3, 1}. If either list runs out, all the nodes should be taken from the other list. The
solution depends on being able to move nodes to the end of a list as discussed in the
Section 1 review. You may want to use MoveNode() as a helper. Overall, many
techniques are possible: dummy node, local reference, or recursion. Using this function
and FrontBackSplit(), you could simulate the shuffling of cards.
/*
Merge the nodes of the two lists into a single list taking a node
alternately from each list, and return the new list.
*/
struct node* ShuffleMerge(struct node* a, struct node* b) {
// Your code
solution:
struct node* ShuffleMerge(struct node* a, struct node* b) {
struct node dummy;
struct node* tail = &dummy;
dummy.next = NULL;
while (1) {
if (a==NULL) { // empty list cases
tail->next = b;
break;
}
else if (b==NULL) {
tail->next = a;
break;
}
else { // common case: move two nodes to tail
tail->next = a;
tail = a;
a = a->next;
tail->next = b;
tail = b;
b = b->next;
}
}
return(dummy.next);
}
between the two lists. So ShuffleMerge() with {1, 2, 3} and {7, 13, 1} should yield {1, 7,
2, 13, 3, 1}. If either list runs out, all the nodes should be taken from the other list. The
solution depends on being able to move nodes to the end of a list as discussed in the
Section 1 review. You may want to use MoveNode() as a helper. Overall, many
techniques are possible: dummy node, local reference, or recursion. Using this function
and FrontBackSplit(), you could simulate the shuffling of cards.
/*
Merge the nodes of the two lists into a single list taking a node
alternately from each list, and return the new list.
*/
struct node* ShuffleMerge(struct node* a, struct node* b) {
// Your code
solution:
struct node* ShuffleMerge(struct node* a, struct node* b) {
struct node dummy;
struct node* tail = &dummy;
dummy.next = NULL;
while (1) {
if (a==NULL) { // empty list cases
tail->next = b;
break;
}
else if (b==NULL) {
tail->next = a;
break;
}
else { // common case: move two nodes to tail
tail->next = a;
tail = a;
a = a->next;
tail->next = b;
tail = b;
b = b->next;
}
}
return(dummy.next);
}