[英]Delete all elements with same value from doubly circular linked list
我有雙重循環鏈表
struct Element {
int data;
Element *next;
Element *previous;
};
struct List {
Element *head;
};
任務是刪除具有相同數據的元素/節點。 例如,如果列表具有1,1,3,4,5,1,並且在解析函數值= 1之后,它應為以下3,4,5。
bool removeAllValue(List &l, int value) {
if(!l.head)
return false;
else if(l.head->next == l.head && l.head->data == value){
l.head = NULL;
return true;
}
Element *p = l.head;
do{
if(p->data == value)
removeElement(l, p);
p = p->next;
}while(p!=l.head);
return true;
上面的函數可以遍歷元素,找到具有給定值/數據的元素,然后調用以下函數刪除元素。
void removeElement(List &l, Element *element){
Element *previous = element->previous;
Element *next = element->next;
Element *head = l.head;
if(next == head){ //if element is tail of the list
l.head->previous = element->previous; // relinking head's previous to new tail
element->previous->next = head;
}else if(element == head){ // if element is head of the list
l.head = next; // relinking head to tail
l.head->previous = previous; // relinking tail to head
}else{
previous->next = next;
next->previous = previous;
}
delete element;
}
我嘗試了不同的方法,但是它總是給我帶來錯誤。 我認為問題在於重新連接尾巴和頭部,但找不到解決方法。
問題出在很少的重新鏈接操作中
void removeElement(List &l, Element *element){
Element *previous = element->previous;
Element *next = element->next;
Element *head = l.head;
if(element->next == element)
l.head = NULL;
else if(next == head){ //if element is tail of the list
Element *newTail = previous;
newTail->next = l.head;
l.head->previous = newTail;
}else if(element == head){ // if element is head of the list
l.head = next; // relinking head
l.head->previous = previous; // relinking tail to head
previous->next = l.head; // relinking head to tail
}else{
previous->next = next;
next->previous = previous;
}
delete element;
}
正如Marek Fekete注意到的那樣,每次刪除元素時,我都沒有更改迭代指針。 因此,每次刪除元素時,我都從頭開始遍歷所有列表。
bool removeAllValue(List &l, int value) {
if(!l.head)
return false;
else if(l.head->next == l.head && l.head->data == value){
l.head = NULL;
return true;
}
Element *p = l.head;
bool end = false;
while(!end){
do{
if(p->data == value){
Element *tmp = p;
p = p->next;
removeElement(l, tmp);
break;
}else
p = p->next;
if(p==l.head || l.head == NULL){
end = true;
break;
}
}while(p);
}
return true;
}
我認為您可以簡化它。 不需要以不同的方式處理removeElement
中的尾部/頭部/中間情況,唯一的好處是在刪除前一個頭部時重置列表的頭部。 我寧願確定列表中的下一個元素和最后一個項目刪除檢測。 像這樣:
Element* removeElement(List &l, Element *element) {
Element *newnext;
Element *previous = element->previous;
Element *next = element->next;
Element *head = l.head;
previous->next = next;
next->previous = previous;
newnext = next;
if (element == head) { // if element is head of the list
l.head = next; // set the head correctly
}
if (element == newnext) // detect if last element in the list was deleted
l.head = newnext = nullptr;
else if (newnext == head) // detect if next is at the head again
newnext = nullptr;
delete element;
return newnext;
}
然后,您不必從頭開始重復刪除每個元素,而不必在特殊情況下處理單個項目列表:
bool removeAllValue(List &l, int value) {
if (!l.head)
return false;
Element *p = l.head;
do {
if (p->data == value)
{
p = removeElement(l, p);
}
else
{
p = p->next;
if (p == l.head)
p = nullptr;
}
} while (p);
return true;
}
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