回文串为首尾对称的字符串：

如a，aba，abba等

单链表思路

1.将字符读入链表

2.找到链表中点

3.将链表从中点断开成2条，将后半条反转

4.比较两条链表是否相等（比较次数以少的为准（长度为奇数时））

双向链表思路

1.将字符读入链表

2.找到链表尾节点

3.从首尾依次向中间比较

（双向链表可以双向移动，代码上更简洁，见下面）

单链表C++代码实现

//// Created by mingm on /3/13.//#include <iostream>#include <math.h>using namespace std;struct Node //节点{char data;Node* next;Node():next(NULL){}Node(char &ch):data(ch),next(NULL){}~Node(){}};class SLinkedList //链表{Node* p_head;public:SLinkedList() //构造函数{p_head = new Node; //带头链表// cout << "new 1" << endl;}~SLinkedList(){ erase(); } //析构函数void erase(){Node *del_tempNode, *tempNode;del_tempNode = p_head;while(del_tempNode != NULL) {tempNode = del_tempNode -> next;delete del_tempNode;// cout << "delete 1" << endl;del_tempNode = tempNode;}}void set_head(Node* p){p_head = p;}Node* get_head(){return p_head;}size_t get_len() //求链表长度{size_t len = 0;Node* p = p_head;while(p){len++;p = p->next;}return len;}void delHeadSentinel() //删除链表表头哨兵{Node* del = p_head;p_head = p_head->next;delete del; //删除链表的表头哨兵// cout << "delete head 1" << endl;}Node* reverse() //链表反转{if(p_head == NULL || p_head->next == NULL)return NULL;else{Node *prevNode, *nextNode, *tempNode;prevNode = p_head;nextNode = prevNode->next;prevNode->next = NULL;while(nextNode != NULL){tempNode = nextNode->next;nextNode->next = prevNode;prevNode = nextNode;nextNode = tempNode;}p_head = prevNode;return p_head;}}Node* findMiddle() //查找链表中点{size_t len = get_len();Node* tempNode = p_head;size_t n = ceil(double(len)/2);for(size_t i = 1; i < n; ++i){tempNode = tempNode->next;}return tempNode;}};int main(){while(true){cout << "-----------------------------------" << endl;char ch;cin.clear();cout << "enter a word, is it a palindrome ?" << endl;if((ch = cin.get()) && ch == '\n'){cout << "empty word !" << endl;continue;}SLinkedList charList, backHalfOfList; //链表(前半部分链表)，后半部分链表Node* tempNode = charList.get_head();while(ch != '\n') //把单词存进链表charList{Node* newNode = new Node(ch);// cout << "new insert 1" << endl;tempNode->next = newNode;tempNode = newNode;ch = cin.get();}charList.delHeadSentinel(); //链表表头删除backHalfOfList.delHeadSentinel(); //把空表头哨兵节点删除Node* endOfFrontList = charList.findMiddle(); //链表的中点是前一半的结束节点Node* backListHead = endOfFrontList->next; //中点的下一个节点是后半部分的开始endOfFrontList->next = NULL;//把前半部分链表断开backHalfOfList.set_head(backListHead); //把后半部分的链表表头地址设置好backHalfOfList.reverse();//后半部分链表反转size_t n = backHalfOfList.get_len(); //求后半部分链表长度Node *frontList = charList.get_head(); //找到前半部分的开头Node *backList = backHalfOfList.get_head(); //后半部分的开头（反转后的）bool answer = false;if(backList == NULL) //如果后半部分为空，说明只有一个字符answer = true;else{for(size_t i = 0; i < n; ++i) //比较数据是否相同{if(frontList->data != backList->data){answer = false;break;}else{answer = true;frontList = frontList->next;backList = backList->next;}}}if(answer)cout << "the word is a palindrome." << endl;elsecout << "the word is not a palindrome." << endl;char conti;cout << "continue to check? (y/n)" << endl;cin >> conti;cin.get();if(conti == 'y' || conti == 'Y'){continue;}elsebreak;}return 0; }

Valgrind检查结果

双向链表C++代码实现

//// Created by mingm on /3/16.//#include <iostream>#include <math.h>using namespace std;struct Node //节点{char data;Node *prev, *next;Node():prev(NULL),next(NULL){}Node(char &ch):data(ch),prev(NULL),next(NULL){}~Node(){}};class SLinkedList //链表{Node* p_head;public:SLinkedList() //构造函数{p_head = new Node; //带头链表// cout << "new 1" << endl;}~SLinkedList(){ erase(); } //析构函数void erase(){Node *del_tempNode, *tempNode;del_tempNode = p_head;while(del_tempNode != NULL) {tempNode = del_tempNode -> next;delete del_tempNode;// cout << "delete 1" << endl;del_tempNode = tempNode;}}void set_head(Node* p){p_head = p;}Node* get_head(){return p_head;}Node* get_tail() //求链表尾节点{Node* p = p_head;if(p_head == NULL)return NULL;while(p->next){p = p->next;}return p;}void delHeadSentinel() //删除链表表头哨兵{Node* del = p_head;p_head = p_head->next;p_head->prev = NULL;delete del; //删除链表的表头哨兵// cout << "delete head 1" << endl;}};int main(){while(true){cout << "-----------------------------------" << endl;char ch;cin.clear();cout << "enter a word, is it a palindrome ?" << endl;if((ch = cin.get()) && ch == '\n'){cout << "empty word !" << endl;continue;}SLinkedList charList; //链表Node* tempNode = charList.get_head();while(ch != '\n') //把单词存进链表charList{Node* newNode = new Node(ch);// cout << "new insert 1" << endl;tempNode->next = newNode; //前面节点后指针指向后面newNode->prev = tempNode; //后面节点前置指针指向前面tempNode = newNode;ch = cin.get();}charList.delHeadSentinel();//链表空表头删除Node *front = charList.get_head(); //定义一个从头开始的指针Node *back = charList.get_tail(); //定义一个从尾部开始的指针bool answer = false;if(front == back) //说明只有一个字符answer = true;else{while(front != back) //比较数据是否相同{if(front->data != back->data){answer = false;break;}else{answer = true;front = front->next;back = back->prev;}}}if(answer)cout << "the word is a palindrome." << endl;elsecout << "the word is not a palindrome." << endl;char conti;cout << "continue to check? (y/n)" << endl;cin >> conti;cin.get();if(conti == 'y' || conti == 'Y'){continue;}elsebreak;}return 0; }

Valgrind检查结果