链表的倒数第k个节点

题目：

求链表的倒数第k个节点

题目要求：

1 输入一个单向链表，输出该链表中倒数第k个结点,

2 链表的倒数第0个结点为链表的尾指针(即与倒数第1个节点)

3 链表为空或者链表结点数小于k，返回空

题目解析：

设置两个指针pre,node; 将pre,node都指向链表第一个节点，然后node向前走pre步，这样pre和node之间就相隔k个节点，然后pre,node同时移动直到node指向链表末尾。此时pre位链表的倒数第k个节点

注意事项：

1 链表是否为空

2 链表节点数小于k

算法实现代码：

//-------------链表、链表结点、链表简单操作定义---------------------struct Node{char c;Node *next;};//链表有头结点,list->head->next指向第一个节点struct List{Node * head;};

//返回list的第k个节点Node* ListBack_K_Node(List *list, unsigned k){//如果链表为空，则返回结果为空if (nullptr == list || nullptr == list->head){return nullptr;}//如果k=0；返回k = 1时的结果if (k == 0){k++;}//node向前走k步Node*node = list->head->next;while ((unsigned)0 != k && nullptr != node){--k;node = node->next;}//链表结点数小于k，则直接返回空if (unsigned(0) != k){return nullptr;}//pre,node同步走，直到node == nullptr,此时pre即为倒数第k个节点Node *pre = list->head->next;while (nullptr != node){node = node->next;pre = pre->next;}return pre;}

链表实现、算法测试代码：

#include <iostream>#include <string>using namespace std;//-------------链表、链表结点、链表简单操作定义---------------------struct Node{char c;Node *next;};//链表有头结点,list->head->next指向第一个节点struct List{Node * head;};Node* AllocNode(){Node *node = new Node;if (nullptr == node){return nullptr;}node->next = nullptr;return node;}void InitList(List *list){list->head = AllocNode();}//在链表头部插入一个结点Node* AddNodeToList(List *list, char c){//如果链表为空，直接返回if (nullptr == list){return nullptr;}//如果链表头结点为空，则对链表初始化if (nullptr == list->head){InitList(list);}Node *node = AllocNode();node->c = c;node->next = list->head->next;list->head->next = node;return list->head->next;}//根据str创建链表Node* CreateList(List *list, const string &str){if (nullptr == list){return nullptr;}for (size_t i = 0; i < str.size(); ++i){AddNodeToList(list, str[i]);}return list->head->next;}//销毁链表void DestroyList(List *list){if (nullptr == list || nullptr == list->head){return;}Node *node = nullptr;while (nullptr != list->head->next){node = list->head->next;list->head->next = node->next;delete node;}delete list->head;list->head = nullptr;}void PrintList(const List* list){if (nullptr == list || nullptr == list->head){return;}for (Node *node = list->head->next; nullptr != node; node = node->next){cout << node->c;}cout << endl;}//---------------------------------------------------------------------------------------------//返回list的第k个节点Node* ListBack_K_Node(List *list, unsigned k){//如果链表为空，则返回结果为空if (nullptr == list || nullptr == list->head){return nullptr;}//如果k=0；返回k = 1时的结果if (k == 0){k++;}//node向前走k步Node*node = list->head->next;while ((unsigned)0 != k && nullptr != node){--k;node = node->next;}//链表结点数小于k，则直接返回空if (unsigned(0) != k){return nullptr;}//pre,node同步走，直到node == nullptr,此时pre即为倒数第k个节点Node *pre = list->head->next;while (nullptr != node){node = node->next;pre = pre->next;}return pre;}void TestListBack_K_Node(List *list, int k){Node *node = ListBack_K_Node(list, k);cout << "Search List's Back " << k << " node ";if (nullptr != node){cout << "success ,node.c = " << node->c << endl;}else{cout << "failure" << endl;}}int main(){string str("123456789");List list;InitList(&list);CreateList(&list, str);PrintList(&list);TestListBack_K_Node(&list, 0);TestListBack_K_Node(&list, 1);TestListBack_K_Node(&list, 5);TestListBack_K_Node(&list, 9);TestListBack_K_Node(&list, 10);TestListBack_K_Node(nullptr, 0);DestroyList(&list);return 0;}

运行结果：