// insert a new data after the given onepublic void insertAfter(int givenData, int newData){ // Your code here Node previous = new Node(givenData); if (previous == null) { //System.out.println("The given previous node cannot be null."); return; } Node newNode = new Node(newData); newNode.next = previous.next; previous.next = newNode;}// Removes the Node with the given datapublic void remove(int current) { // Your code here Node previous = new Node(current); int count = 1; int position = 0; while (count < position -1) { previous = previous.next; count++; } Node curNode = previous.next; previous.next = curNode.next; curNode.next = null;}此代碼沒有給出任何錯誤消息，但不會添加或刪除節點。我相信我的困惑在于int givenData以及如果它不是節點如何訪問它。第一次在這里發帖，希望我提供了足夠的信息:)public class LinkedListTemplate {Node head;// inserts data to the end of the list only using the head pointerpublic void append(int data){ Node newNode = new Node(data); if(head == null){ head = newNode; } else { Node currentNode = head; while(currentNode.next != null){ currentNode = currentNode.next; } currentNode.next = newNode; }}// inserts data to the beginning of the listpublic void prepend(int data){ if(head == null){ Node newNode = new Node(data); head = newNode; return; } Node newNode = new Node(data); newNode.next = head; head = newNode;}// print the linked list elementspublic void print() { Node currentNode = head; System.out.printf("["); while (currentNode.next != null) { System.out.printf("%d, ", currentNode.data); currentNode = currentNode.next; } System.out.printf("%d]%n", currentNode.data);}我包含了其余的鏈表代碼，還有另一個類可以測試列表并允許它運行。
3 回答
翻翻過去那場雪
TA貢獻2065條經驗 獲得超14個贊
將插入方法更改為：
public void insertAfter(int givenData, int newData) {
Node n = head;
while (n != null && n.data != givenData) {
n = n.next;
}
if (n != null) return;
Node newNode = new Node(newData);
newNode.next = n.next;
n.next = newNode;
}
并使用以下方法刪除方法：
public void remove(int current) {
Node prev = head;
if(head == null) return;
if(head.data == current){head = null; return;}
Node n = prev.next;
while( n != null && n.data != current){
prev = n;
n = n.next;
}
if (n == null) return;
prev.next = n.next;
}
慕的地8271018
TA貢獻1796條經驗 獲得超4個贊
好的，您可能需要稍微增強您的代碼。首先，你需要一個清單。所以讓我們從那個開始。
public class MyLinkedList {
public Node *head = NULL;
}
head 指向列表的前面。
現在， insertAfter 的基本算法很簡單：
查找包含“插入位置”值的節點
創建一個新節點并在之后插入。
可愛吧？
void insertAfter(int beforeData, int valueToInsert) {
Node *ptr = head;
Node *lastPtr = NULL;
while (ptr != NULL && ptr->value != beforeData) {
lastPtr = ptr;
ptr = ptr->next;
}
// At this point, ptr COULD be null, if the data is never found.
// Insertion is at the end, and lastPtr helps with that.
Node * newNode = new Node(valueToInsert);
if (ptr == NULL) {
# At the end of the list.
if (lastPtr != NULL) {
lastPtr->next = newNode;
newNode->previous = lastPtr;
}
else {
// The entire list is null
head = newNode;
}
}
else {
// ptr points to the data we were searching for, so insert after.
newNode->next = ptr->next;
newNode->previous = ptr;
ptr->next = newNode;
}
}
然后，您可以執行以下操作：
MyLinkedList list;
list.insertAfter(0, 1);
list.insertAfter(1, 17);
list.insertAfter(1, 10);
如果您隨后轉儲列表的內容（您需要從 head 遍歷 next 指針），您將看到列表按 1、10、17 的順序包含。
富國滬深
TA貢獻1790條經驗 獲得超9個贊
我試了一下。
請注意，我實際上并不知道 remove 真正應該做什么。
您的代碼正在尋找具有給定索引的節點，因此我堅持使用該節點（第一個刪除版本）。
但評論建議，您需要刪除具有給定數據的節點。所以我添加了第二個版本，就是這樣做的（第二個刪除版本）。
public void insertAfter(int givenData, int newData) {
// Your code here
Node previous = head;
while (null != previous && previous.data != givenData) {
previous = previous.next;
}
// this prevents insertion if given Data was not found
if (previous == null) {
System.out.println("the given data doesn't exist");
return;
}
Node newNode = new Node(newData);
newNode.next = previous.next;
previous.next = newNode;
}
刪除具有給定索引的節點的版本：
// Removes the Node with the given index
public void remove(int current) {
Node previous = head;
for (int i = 0; i < current - 1; i++) {
if (null == previous) {
System.out.println("the given position doesn't exist");
return;
}
previous = previous.next;
}
if (current == 0) {
head = head.next;
} else if (previous.next == null) {
System.out.println("the given position is after the end - nothing to do!");
} else {
previous.next = previous.next.next;
}
}
刪除具有給定數據的元素的版本
// Removes the Node with the given data
public void remove(int data) {
Node previous = head;
if(head == null){
System.out.println("Empty list - nothing to remove!");
return;
} if (head.data == data){
head = head.next;
return;
}
while(previous.next != null && previous.next.data != data){
previous = previous.next;
}
if (previous.next == null) {
System.out.println("the given element was not found - nothing to do!");
} else {
previous.next = previous.next.next;
}
}
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