我遇到了问题,我正在尝试创建一个删除最高值保留号的列表,或者如果列表中的值最高则创建具有相同值的所有数字。感谢您提供任何建议。
// n,n1,head,next - are pointers
int j = 0; //this number helps to put pointer forward by one place
while(n!=0){//should go through every digit of the list
if(head == 0){
cout << "list is empty" << endl;
}
else{
n = head;
n1=0; // n1 and n are pointers
while(n!=0){
if(n->sk == maxx){//searches for maximum digit in the list
break;
}
else{
n1=n;
n=n->next;
}
}
if(head == n){
head = head->next;
}
else{
n1->next = n->next;
}
delete n; // deletes the pointer holding the highest value
}
n = head; //problem is here or somewhere below
j++;
for(int i=0; i<j;i++){ // this loop should make the pointer point to the first
n = n->next; // number, then the second and so on until the end of list
} // and all the numbers inside the list with the value that
} // equals "maxx" should be deleted
答案 0 :(得分:2)
你应该取消引用指针。现在,你指着他们的地址。看看这是否有助于解决您的问题。
答案 1 :(得分:0)
好的,问题(大部分)是代码:
while(n!=0){
if(n->sk == maxx){
break;
}
else{
n1=n;
n=n->next;
}
}
如果找到maxx
值,则应删除该节点并继续搜索,不要break
。这样,您就不需要这么多代码来完成这项任务。
while (n != 0){
if (n->sk == maxx){
node *prev = n->prev; // The previous node.
node *tmp = n; // this assume you have a class node.
// temporaly holds the pointer to n.
prev->next = n->next; // Connect the previous node with the following one.
n = n->next; // advance n to the next node in the list.
delete tmp; // delete the node.
}
}
答案 2 :(得分:0)
如果我理解你想要什么,你可以迭代你的列表并保存指针以便删除:
it = head;
pos = nullptr;
while (it != nullptr) {
if(it -> sk == maxx) {
pos = it; // save ptr
it = it -> next;
delete pos; // delete saved ptr
pos = nullptr;
}
}