[英]Reversing a singly-linked list in C++
我試圖以遞歸方式反轉單鏈表的實現時遇到問題。
我已經閱讀了關於這個過程的其他類似問題,但是,在我嘗試在我自己的程序中實現這個過程時,我做得很簡短。
這是我在下面的嘗試(與之后的代碼中顯示的略有不同):
注意:我的列表使用root
指針,該指針不包含重要數據,僅用作引用列表中數據的地址。
void IntList::reverse(Node* t_node) {
if(t_node == NULL) {
reverse(root);
} else
if(t_node->next == NULL) {
cout << "In (swapping): " << t_node->value << endl;
root->next = t_node;
} else {
cout << "In: " << t_node->value << endl;
Node* tmp = t_node->next;
reverse(t_node->next);
tmp->next = t_node;
}
return NULL;
}
我在某個地方丟失了引用,並在嘗試顯示列表時無休止地打印。 我真的不知道我犯的錯誤是什么,但懷疑它可能與我如何處理我的root
。
這是完整程序的全部(除了reverse()
方法之外的所有功能)。
#ifndef INTLIST_H
#define INTLIST_H
#include<iostream>
using namespace std;
class IntList {
private:
struct Node {
int value;
Node* next;
};
int size;
Node* root;
void destroy();
public:
IntList() { root = new Node; root->next = 0; root-> value = 0; size = 0;}
IntList(const IntList& list) { this->root = list.root; this->size = list.size; }
~IntList() {}
void appendNode(int val);
void insertNode(int pos, int val);
void deleteNode(int pos);
int searchNode(int val);
int getSize() const;
void print() const;
Node* reverse(Node* t_node);
int &operator[](int element) const;
void pop_back();
void pop_front();
void push_back(int val);
void push_front(int val);
};
void IntList::appendNode(int val) {
push_back(val);
}
void IntList::insertNode(int pos, int val) {
Node* tmp;
Node* current = root;
for(int i = 0; i < pos && current->next != NULL; i++) {
current = current->next;
}
tmp = new Node;
tmp->value = val;
tmp->next = current->next;
current->next = tmp;
size++;
}
void IntList::deleteNode(int pos) {
Node* tmp;
Node* current = root;
if(pos <= size-1) {
for(int i = 0; i < pos; i++) {
current = current->next;
}
tmp = current->next;
current->next = tmp->next;
delete tmp;
size--;
} else {
cout << "ERROR: Out of range." << endl;
}
}
int IntList::searchNode(int val) {
int position = 0;
Node* current = root->next;
if(size != 0) {
for(position = 0; position < size && current->value != val; position++) {
current = current->next;
}
} else {
cout << "ERROR: List is empty." << endl;
position = -1;
}
return position;
}
int IntList::getSize() const {
return size;
}
void IntList::print() const {
Node* current = root->next;
cout << "List: ";
while(current != NULL) {
cout << current->value << " ";
current = current->next;
}
if(getSize() == 0) {
cout << "Empty.";
}
cout << endl;
}
IntList::Node* IntList::reverse(Node* t_node) {
#define REVERSE
#ifndef REVERSE
if(t_node == NULL) {
reverse(root);
} else
if(t_node->next == NULL) {
cout << "In (swapping): " << t_node->value << endl;
root->next = t_node;
} else {
cout << "In: " << t_node->value << endl;
Node* tmp = t_node->next;
reverse(t_node->next);
tmp->next = t_node;
}
#endif //reverses list, but causes infinite loop in display
return NULL;
}
int &IntList::operator[](int pos) const {
Node* current = root->next;
if(pos <= size-1) {
for(int i = 0; i < pos; i++) {
current = current->next;
}
} else {
cout << "ERROR: Out of bounds.";
current = NULL;
}
return current->value;
}
void IntList::pop_back() {
deleteNode(size-1);
}
void IntList::pop_front() {
deleteNode(0);
}
void IntList::push_back(int val) {
insertNode(size, val);
}
void IntList::push_front(int val) {
insertNode(0, val);
}
#endif
#ifndef LINKEDLIST_H
#define LINKEDLIST_H
#include<iostream>
using namespace std;
template<typename T>
class LinkedList {
private:
struct Node {
T value;
Node* next;
};
int size;
Node* root;
void destroy();
public:
LinkedList() { root = new Node; root->next = 0; root-> value = 0; size = 0;}
LinkedList(const LinkedList &) {}
~LinkedList() {}
void appendNode(T val);
void insertNode(int pos, T val);
void deleteNode(int pos);
int searchNode(T val);
int getSize() const;
void print() const;
void reverse(Node* t_node);
int &operator[](int element) const;
void pop_back();
void pop_front();
void push_back(T val);
void push_front(T val);
};
template <typename T>
void LinkedList<T>::appendNode(T val) {
push_back(val);
}
template <typename T>
void LinkedList<T>::insertNode(int pos, T val) {
Node* tmp;
Node* current = root;
for(int i = 0; i < pos && current->next != NULL; i++) {
current = current->next;
}
tmp = new Node;
tmp->value = val;
tmp->next = current->next;
current->next = tmp;
size++;
}
template <typename T>
void LinkedList<T>::deleteNode(int pos) {
Node* tmp;
Node* current = root;
if(pos <= size-1) {
for(int i = 0; i < pos; i++) {
current = current->next;
}
tmp = current->next;
current->next = tmp->next;
delete tmp;
size--;
} else {
cout << "ERROR: Out of range." << endl;
}
}
template <typename T>
int LinkedList<T>::searchNode(T val) {
int position = 0;
Node* current = root->next;
if(size != 0) {
for(position = 0; position < size && current->value != val; position++) {
current = current->next;
}
} else {
cout << "ERROR: List is empty." << endl;
position = -1;
}
return position;
}
template <typename T>
int LinkedList<T>::getSize() const {
return size;
}
template <typename T>
void LinkedList<T>::print() const {
Node* current = root->next;
cout << "List: ";
while(current != NULL) {
cout << current->value << " ";
current = current->next;
}
if(getSize() == 0) {
cout << "Empty.";
}
cout << endl;
}
template <typename T>
void LinkedList<T>::reverse(Node* t_node) {
/*
if(t_node == NULL) {
reverse(root);
} else
if(t_node->next == NULL) {
cout << "In (swapping): " << t_node->value << endl;
root->next = t_node;
} else {
cout << "In: " << t_node->value << endl;
Node* tmp = t_node->next;
reverse(t_node->next);
tmp->next = t_node;
}
*/ //reverses list, but causes infinite loop in display
}
template <typename T>
int &LinkedList<T>::operator[](int pos) const {
Node* current = root->next;
if(pos <= size-1) {
for(int i = 0; i < pos; i++) {
current = current->next;
}
} else {
cout << "ERROR: Out of bounds.";
current = NULL;
}
return current->value;
}
template <typename T>
void LinkedList<T>::pop_back() {
deleteNode(size-1);
}
template <typename T>
void LinkedList<T>::pop_front() {
deleteNode(0);
}
template <typename T>
void LinkedList<T>::push_back(T val) {
insertNode(size, val);
}
template <typename T>
void LinkedList<T>::push_front(T val) {
insertNode(0, val);
}
#endif
//test driver
int main() {
IntList i_list;
int n;
cout << "Appending node: value = " << 0 << endl;
i_list.appendNode(0);
i_list.print();
cout << endl;
n = 5;
cout << "Inserting nodes (at their values). Node values = { ";
for(int i = 0; i < n; i++) {
cout << i << " ";
i_list.insertNode(i,i);
}
cout << "}" << endl;
i_list.print();
cout << endl;
cout << "Deleting node at position: " << i_list.getSize()-1 << endl;
i_list.deleteNode(i_list.getSize()-1);
i_list.print();
cout << endl;
cout << "Searching for value: " << 3 << endl;
cout << "Found at: " << i_list.searchNode(3) << endl;
cout << endl;
i_list.print();
cout << "List size: " << i_list.getSize() << endl;
cout << endl;
n = 3;
cout << "Calling node at list[" << n << "]: " << i_list[n] << endl;
cout << endl;
i_list.print();
cout << "Deleting node from back position." << endl;
i_list.pop_back();
i_list.print();
cout << endl;
i_list.print();
cout << "Deleting node from front position." << endl;
i_list.pop_front();
i_list.print();
cout << endl;
n = 9;
i_list.print();
cout << "Adding node (value = " << n << ") to back position." << endl;
i_list.push_back(n);
i_list.print();
cout << endl;
n = 8;
i_list.print();
cout << "Adding node (value = " << n << ") to front position." << endl;
i_list.push_front(n);
i_list.print();
cout << endl;
i_list.print();
cout << "Copying list to new list." << endl;
IntList t_list(i_list);
cout << endl;
cout << "List copy:" << endl;
t_list.print();
cout << endl;
/*
* Showing functionality transfers over to LinkedList template class
* generally, for primitive data types (lacks absolutely generality
* for data which can't be passed directly to cout).
*/
cout << "List functionality transfers generally to LinkedList class:" << endl;
LinkedList<int> int_list;
LinkedList<double> double_list;
LinkedList<char> char_list;
cout << "Appending nodes:" << endl;
n = 5;
for(int i = 0; i < n; i++){
int_list.appendNode(i);
}
int_list.print();
n = 5;
for(int i = 0; i < n; i++){
double_list.appendNode(i+0.1);
}
double_list.print();
n = 5;
for(int i = 0; i < n; i++){
char_list.appendNode('A' + i);
}
char_list.print();
cout << "Removing nodes:" << endl;
n = 5;
for(int i = 0; i < n; i++){
int_list.pop_back();
}
int_list.print();
n = 5;
for(int i = 0; i < n; i++){
double_list.pop_back();
}
double_list.print();
n = 5;
for(int i = 0; i < n; i++){
char_list.pop_back();
}
char_list.print();
return 0;
}
編輯 :我修改了我的算法,我相信它在算法上有效,但在功能上它可能正在做一些導致內存問題的事情。 我不確定為什么會這樣,但這里是:
void IntList::reverse() {
IntList tmp(*this);
int list_size = size;
for(int i = 0; i < list_size; i++) {
this->insertNode(i, tmp[tmp.getSize()-1]);
this->pop_back();
tmp.pop_back();
}
}
事實上,如果我的[]
運算符重載在這個方法中運行(由於某種原因它不是?)我可以取消tmp
列表並直接引用列表中的最后一個值作為this[size-1]
。
這是什么問題?
你的問題是在reverse()
之后,列表中的最后一個元素將指向根元素而不是指向null。 一種解決方案可能是明確檢查該條件,以便您獲得:
void IntList::reverse(Node* t_node) {
if(t_node == NULL) {
reverse(root);
return;
}
if(t_node->next == NULL) {
cout << "In (swapping): " << t_node->value << endl;
root->next = t_node;
} else {
cout << "In: " << t_node->value << endl;
Node* tmp = t_node->next;
reverse(t_node->next);
// If this node was the first node it will now be the last
if (t_node == root) {
tmp->next = NULL;
} else {
tmp->next = t_node;
}
}
}
如果應該可以反轉列表的子部分,那么這不起作用。 如果這是您需要的東西,那么您可能需要使用輔助函數來處理除第一個元素之外的所有元素。
假設我們有IntList
{1,2,3},它實際上有這種形式:
0 -> 1 -> 2 -> 3
然后我們調用reverse(root)
,這樣t_node
與root
具有相同的值(因此指向(0))。
Node* tmp = t_node->next;
所以tmp指向(1)。
reverse(t_node->next);
假設這是有效的,列表現在是0-> 1-> 3-> 2
tmp->next = t_node;
所以現在1-> 0。 該列表現在是一個循環,其他節點已丟失。
目前尚不清楚你打算做什么這個功能,但你必須誤解一些東西。
編輯:當您不了解低級別的機制時,您正在嘗試高級解決方案。
你的副本構造函數:
IntList(const IntList& list) { this->root = list.root; this->size = list.size; }
執行我們所謂的“淺拷貝”; 它復制指針成員,但不復制他們指向的東西。 如果您有一個如下所示的列表A
:
0->1->2->3
然后調用IntList B(A);
,你會得到這樣的東西:
0
|
v
0->1->2->3
如果你然后調用A.pop_back()
和B.pop_back()
, 你認為會發生什么?
更重要的是,你想做什么? 你想知道如何編寫遞歸函數,還是不再需要?
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