VietMX Staff asked 3 years ago
The add() method below walks down the list until it finds the appropriate position. Then, it splices in the new node and updates the start, prev, and curr pointers where applicable.
Note that the reverse operation, namely removing elements, doesn’t need to change, because you are simply throwing things away which would not change any order in the list.
Implementation
public void add(T x) {
Node newNode = new Node();
newNode.info = x;
// case: start is null; just assign start to the new node and return
if (start == null) {
start = newNode;
curr = start;
// prev is null, hence not formally assigned here
return;
}
// case: new node to be inserted comes before the current start;
// in this case, point the new node to start, update pointers, and return
if (x.compareTo(start.info) < 0) {
newNode.link = start;
start = newNode;
curr = start;
// again we leave prev undefined, as it is null
return;
}
// otherwise walk down the list until reaching either the end of the list
// or the first position whose element is greater than the node to be
// inserted; then insert the node and update the pointers
prev = start;
curr = start;
while (curr != null && x.compareTo(curr.info) >= 0) {
prev = curr;
curr = curr.link;
}
// splice in the new node and update the curr pointer (prev already correct)
newNode.link = prev.link;
prev.link = newNode;
curr = newNode;
}