[英]Remove Duplicates linked list
void RemoveDuplicates(Slist& l)
{
if (l.head == NULL) {
return;
}
Node* cur = l.head;
while (cur != NULL && cur->next != NULL) {
Node* prev = cur;
Node* temp = cur->next;
while (temp != NULL) {
if (temp->data == cur->data) {
prev->next = temp->next;
cur->next = prev->next;
temp = prev->next;
}
else {
prev = prev->next;
temp = temp->next;
}
}
cur = cur->next;
}
}
嗨,我想从链表中删除重复项(0 为 NULL)
input: 1->2->2->4->2->6->0
outPut: 1->2->4->6->0
我运行程序后的结果是:
1->2->6
我哪里错了? 请帮我
这是我对您的问题的解决方案:
bool alreadyExist(Node head)
{
Node cur = head;
while(cur.next != nullptr)
{
if(cur.next->data == head.data) {
return true;
}
cur = *cur.next;
}
return false;
}
void RemoveDuplicates(Slist& l)
{
if (l.head == nullptr) {
return;
}
Node* head = l.head;
Node* curPtr = l.head->next;
while(curPtr != nullptr)
{
if(alreadyExist(*curPtr) == false)
{
head->next = curPtr;
head->next->prev = head;
head = head->next;
curPtr = curPtr->next;
}
else
{
Node* backup = curPtr;
curPtr = curPtr->next;
// delete duplicate elements from the heap,
// if the nodes were allocated with new, malloc or something else
// to avoid memory leak. Remove this, if no memory was allocated
delete backup;
}
}
}
重要提示:节点对象的析构函数不允许删除 next 和 prev 指针后面的链接 object。
对于您的输入示例,它会导致 output 1->4->2->6->0
。 它的顺序并不完全准确,您想要 output,但每个数字在 output 中只存在一次。 它只添加重复号码的最后一次。 I don't really know, if you use C or C++, but because I prefer C++, I replaced the NULL with nullptr in the code. 如果对象不在使用 malloc 或新创建的 HEAP 上,则可以删除删除。
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