【c++版数据结构】单链表复习之常见面试题型1

题目描述:

1：从尾到头依次打印单链表

2：在链表中删除指定的节点

3：假设有一个没有头指针的单链表。一个指针指向此单链表中间的一个节点（不是第一个，也不是最后一个节点），请将该节点从单链表中删除。

4：假设有一个没有头指针的单链表,一个指针指向此单链表中一个节点,要求在此节点前插入一个节点

5：编写一个函数，给定一个链表的头指针，要求只遍历一次，将单链表中的元素顺序反转过来

解题思路及其代码如下：（单链表采用模板类，无头节点）

List.h

#ifndef _LIST_H_#define _LIST_H_#include<iostream>#include<cstdlib>#include<cassert>#include<stack>using namespace std;template<typename T>class List;template<typename T>class ListNode{friend class List<T>;public:ListNode(const T val = 0, ListNode<T> *next = NULL) :_val(val), _next(NULL){}private:T _val;ListNode<T> *_next;};template<typename T>class List{public:List(){this->head = NULL;}bool push_back(const T&val){ListNode<T> *new_node = buy_node(val, NULL);if (NULL == new_node){return false;}if (head == NULL){head = new_node;}else{ListNode<T> *tmp = head;while (NULL != tmp->_next){tmp = tmp->_next;}tmp->_next = new_node;}return true;}bool push_front(const T&val){ListNode<T> *new_node = buy_node(val, NULL);if (NULL == new_node){return false;}else{new_node->_next = head;head = new_node;}return true;}bool pop_back(){if (is_empty()){cout << "the list is already empty,can not be pop!" << endl;return false;}if (head->_next == NULL){delete head;head = NULL;}else{ListNode<T> *tmp = head;while (tmp->_next->_next != NULL){tmp = tmp->_next;}delete tmp->_next;tmp->_next = NULL;}return true;}bool pop_front(){if (is_empty()){cout << "the list is already empty,can not be pop!" << endl;return false;}else{ListNode<T> *tmp = head;head = head->_next;delete tmp;return true;}}~List(){while (head != NULL){pop_back();}}void print()const{ListNode<T> *tmp = head;while (NULL != tmp){cout << tmp->_val << "->";tmp = tmp->_next;}cout <<"NULL"<< endl;}//<<从尾到头依次打印单链表>>//1：栈 2:递归//如果链表元素很多，递归的方式会导致栈溢出，所以第一种方式鲁棒性更好void print_from_tail_by_stack()const{stack<ListNode<T> *> st;ListNode<T> *tmp = head;while (tmp != NULL){st.push(tmp);tmp = tmp->_next;}while (!st.empty()){tmp = st.top();        //注意，stack的pop函数只是将栈顶元素删除st.pop();              //（无法获取栈顶元素）->若要获取必须要通过top函数cout << tmp->_val << "->";}cout << "NULL" << endl;}void print_form_tail_by_recursion(){_print_form_tail_by_recursion(head);cout <<"NULL"<< endl;}//在链表中删除指定的节点bool delete_node(const T &val){if (head->_val == val){ListNode<T> *tmp = head;head = head->_next;delete tmp;}else{ListNode<T> *prev = head;while (prev->_next != NULL && prev->_next->_val != val){prev = prev->_next;}if (prev->_next == NULL){cout << "it is not exist!" << endl;return false;}else{ListNode<T> *tmp = prev->_next;prev->_next = tmp->_next;delete tmp;}}return true;}//问题描述：假设有一个没有头指针的单链表。一个指针指向此单链表中间的//一个节点（不是第一个，也不是最后一个节点），请将该节点从单链表中删除。//解题思路：将该节点的后面一个节点的内容，拷贝到该节点，然后将后一个节点删除void delete_mid_node(ListNode<T> *cur){ListNode<T> *next = cur->_next;cur->_val = next->_val;cur->_next = next->_next;delete next;}//编写一个函数，给定一个链表的头指针，要求只遍历一次，将单链表中的元素顺序反转过来。//指向第二个节点，然后将第二个节点以及之后的节点头插void reverse1(){if (head == NULL || head->_next == NULL){//链表为空或者只有一个节点，无需逆置return;}ListNode<T> *tmp = head->_next;head->_next = NULL;while (tmp != NULL){ListNode<T> *new_head = tmp;tmp = tmp->_next;new_head->_next = head;head = new_head;}}void reverse2(){//三个指针if (head == NULL || head->_next == NULL){//链表为空或者只有一个节点，无需逆置return;}ListNode<T> *prev = head;ListNode<T> *cur = prev->_next;ListNode<T> *next = NULL;head->_next = NULL;while (cur != NULL){next = cur->_next;cur->_next = prev;head = cur;prev = cur;cur = next;}}//问题描述：假设有一个没有头指针的单链表,一个指针指向此单链表中一个节点,要求在//此节点前插入一个节点//思路：在该节点之后插入该节点，然后交换二者的数据void insert(ListNode<T> *cur,int val){assert(cur != NULL);ListNode<T> *new_node = buy_node(val, NULL);assert(new_node != NULL);new_node->_next = cur->_next;cur->_next = new_node;swap(cur->_val, new_node->_val);}private:ListNode<T> *buy_node(const T val = 0, ListNode<T> *next = NULL){ListNode<T> *tmp = new ListNode<T>(val,next);return tmp;}bool is_empty()const{return head == NULL;}void _print_form_tail_by_recursion(ListNode<T> *head){//递归方式if (head == NULL){return;}else{_print_form_tail_by_recursion(head->_next);cout << head->_val << "->";}}ListNode<T> *head;};#endif

main.cpp

#include"List.h"void test1(){List<int> list;for (int i = 0; i < 10; ++i){list.push_front(i);}list.print();//list.print_form_tail_by_recursion();//list.print_from_tail_by_stack();//for (int i = 0; i < 5; ++i){//list.pop_back();//list.pop_front();//}//list.print();}//void test2(){List<int> list;for (int i = 0; i < 10; ++i){list.push_front(i);}list.print();list.delete_node(0);list.print();list.delete_node(3);list.print();list.delete_node(9);list.print();}//reversevoid test3(){List<int> list;for (int i = 0; i < 10; ++i){list.push_front(i);}list.print();list.reverse2();//list.reverse1();list.print();}int main(){test3();system("pause");return 0;}