回文串為首尾對稱的字符串：

如a，aba，abba等

單鏈表思路

1.將字符讀入鏈表

2.找到鏈表中點

3.將鏈表從中點斷開成2條，將后半條反轉

4.比較兩條鏈表是否相等（比較次數以少的為準（長度為奇數時））

雙向鏈表思路

1.將字符讀入鏈表

2.找到鏈表尾節點

3.從首尾依次向中間比較

（雙向鏈表可以雙向移動，代碼上更簡潔，見下面）

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單鏈表C++代碼實現

// // Created by mingm on 2019/3/13. // #include <iostream> #include <math.h> using namespace std; struct Node //節點 {char data;Node* next;Node():next(NULL){}Node(char &ch):data(ch),next(NULL){}~Node(){} }; class SLinkedList //鏈表 {Node* p_head; public:SLinkedList() //構造函數{p_head = new Node; //帶頭鏈表 // cout << "new 1" << endl;}~SLinkedList(){ erase(); } //析構函數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() //求鏈表長度{size_t len = 0;Node* p = p_head;while(p){len++;p = p->next;}return len;}void delHeadSentinel() //刪除鏈表表頭哨兵{Node* del = p_head;p_head = p_head->next;delete del; //刪除鏈表的表頭哨兵 // cout << "delete head 1" << endl;}Node* reverse() //鏈表反轉{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() //查找鏈表中點{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; //鏈表(前半部分鏈表)，后半部分鏈表Node* tempNode = charList.get_head();while(ch != '\n') //把單詞存進鏈表charList{Node* newNode = new Node(ch); // cout << "new insert 1" << endl;tempNode->next = newNode;tempNode = newNode;ch = cin.get();}charList.delHeadSentinel(); //鏈表表頭刪除backHalfOfList.delHeadSentinel(); //把空表頭哨兵節點刪除Node* endOfFrontList = charList.findMiddle(); //鏈表的中點是前一半的結束節點Node* backListHead = endOfFrontList->next; //中點的下一個節點是后半部分的開始endOfFrontList->next = NULL; //把前半部分鏈表斷開backHalfOfList.set_head(backListHead); //把后半部分的鏈表表頭地址設置好backHalfOfList.reverse(); //后半部分鏈表反轉size_t n = backHalfOfList.get_len(); //求后半部分鏈表長度Node *frontList = charList.get_head(); //找到前半部分的開頭Node *backList = backHalfOfList.get_head(); //后半部分的開頭（反轉后的）bool answer = false;if(backList == NULL) //如果后半部分為空，說明只有一個字符answer = true;else{for(size_t i = 0; i < n; ++i) //比較數據是否相同{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檢查結果

雙向鏈表C++代碼實現

// // Created by mingm on 2019/3/16. // #include <iostream> #include <math.h> using namespace std; struct Node //節點 {char data;Node *prev, *next;Node():prev(NULL),next(NULL){}Node(char &ch):data(ch),prev(NULL),next(NULL){}~Node(){} }; class SLinkedList //鏈表 {Node* p_head; public:SLinkedList() //構造函數{p_head = new Node; //帶頭鏈表 // cout << "new 1" << endl;}~SLinkedList(){ erase(); } //析構函數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() //求鏈表尾節點{Node* p = p_head;if(p_head == NULL)return NULL;while(p->next){p = p->next;}return p;}void delHeadSentinel() //刪除鏈表表頭哨兵{Node* del = p_head;p_head = p_head->next;p_head->prev = NULL;delete del; //刪除鏈表的表頭哨兵 // 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; //鏈表Node* tempNode = charList.get_head();while(ch != '\n') //把單詞存進鏈表charList{Node* newNode = new Node(ch); // cout << "new insert 1" << endl;tempNode->next = newNode; //前面節點后指針指向后面newNode->prev = tempNode; //后面節點前置指針指向前面tempNode = newNode;ch = cin.get();}charList.delHeadSentinel(); //鏈表空表頭刪除Node *front = charList.get_head(); //定義一個從頭開始的指針Node *back = charList.get_tail(); //定義一個從尾部開始的指針bool answer = false;if(front == back) //說明只有一個字符answer = true;else{while(front != back) //比較數據是否相同{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檢查結果?

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