练手小程序2.链表翻转问题

来源：互联网发布：上海希格斯网络金融编辑：程序博客网时间：2024/06/07 04:09

问题1：给出一个链表，将其翻转，比如链表1→2→3→4→5→6，翻转后6→5→4→3→2→1；
问题2：给出一个链表和一个数k，将链表进行翻转，比如链表1→2→3→4→5→6，k=2，

则翻转后2→1→4→3→6→5，若k=3,翻转后3→2→1→6→5→4，若k=4，翻转后4→3→2→1→5→6

考察的知识点：链表的灵活操作

代码如下：

#include "stdafx.h"#include <iostream>#include <vector>//链表结点类型struct List_Node{int m_num_;List_Node *m_next_;};//尾插法创建一个单链表void creat_list_from_tail(List_Node *&p_head){std::cout << "尾插法创建一个单链表" << std::endl;while (1){//新建一个结点List_Node *new_node = new List_Node;std::cout << "please input a number: ";std::cin >> new_node->m_num_;new_node->m_next_ = NULL;if (-1 == new_node->m_num_){return;}if (NULL == p_head)//链表为空{p_head = new_node;}else{List_Node *p_cur = p_head;while (p_cur->m_next_ != NULL)//找到最后一个结点{p_cur = p_cur->m_next_;}p_cur->m_next_ = new_node;}}}//头插法创建一个单链表void creat_list_from_head(List_Node *&p_head){std::cout << "头插法创建一个单链表" << std::endl;while (1){//新建一个结点List_Node *new_node = new List_Node;std::cout << "please input a number: ";std::cin >> new_node->m_num_;new_node->m_next_ = NULL;if (-1 == new_node->m_num_){return;}if (NULL == p_head)//链表为空{p_head = new_node;}else{new_node->m_next_ = p_head;p_head = new_node;}}}//打印链表void print_list(List_Node *p_head){if (NULL == p_head){std::cout << "链表为空！" << std::endl;return;}List_Node *p_cur = p_head;std::cout << "链表中的元素为：";while (NULL != p_cur){std::cout << p_cur->m_num_ << " ";    p_cur = p_cur->m_next_;}std::cout << std::endl;}//翻转链表void reverse_list(List_Node *&list_head){std::cout << "翻转链表！" << std::endl;if (list_head == NULL){std::cout << "链表为空！" << std::endl;return;}List_Node *cur_node = list_head;//当前结点List_Node *pre_node = NULL;//当前结点的前一个结点List_Node *next_node = NULL;//当前结点的下一个结点    while (cur_node != NULL)    {next_node = cur_node->m_next_;//记录当前结点的下一个结点        cur_node->m_next_ = pre_node;pre_node = cur_node;cur_node = next_node;    }list_head = pre_node;}//将链表list_head按每k个元素切割成若干小的链表存放到list_vec中，且链表内部实现了翻转void split_list(List_Node *&list_head, int &k, std::vector<List_Node *> &list_vec){List_Node *cur_node = list_head;//当前结点List_Node *pre_node = NULL;//当前结点的前一个结点List_Node *next_node = NULL;//当前结点的下一个结点List_Node *list_tmp = NULL;int count = k;while (cur_node != NULL && count > 0){--count;next_node = cur_node->m_next_;//记录当前结点的下一个结点cur_node->m_next_ = pre_node;pre_node = cur_node;cur_node = next_node;if (count == 0 || cur_node == NULL){count = k;list_vec.push_back(pre_node);pre_node = NULL;}}}//将若干小的链表链接成一个大的链表void connect_list_vec(std::vector<List_Node *> list_vec, List_Node *&list_head){List_Node *cur_tail = NULL;bool is_first = true;for (std::vector<List_Node *>::iterator iter = list_vec.begin();iter != list_vec.end(); ++iter){if (is_first){list_head = *iter;cur_tail = *iter;is_first = false;}else{cur_tail->m_next_ = *iter;}while (cur_tail->m_next_ != NULL){cur_tail = cur_tail->m_next_;}}}//翻转链表,指定每K个元素翻转一次void reverse_list(List_Node *list_head, int &k){if (list_head == NULL){std::cout << "链表为空！" << std::endl;return;}std::cout << "翻转链表！" << std::endl;while (1){std::cout << "请输入一个数k: ";std::cin >> k;if (-1 == k){return;}std::vector<List_Node *> list_vec;split_list(list_head, k, list_vec);connect_list_vec(list_vec, list_head);print_list(list_head);}}int _tmain(int argc, _TCHAR* argv[]){List_Node *list_head = NULL;creat_list_from_tail(list_head);//创建链表print_list(list_head);//打印链表int k = 0;reverse_list(list_head, k);//翻转链表system("pause");return 0;}

运行结果：

0 0