利用栈和队列完成回文的判断代码

好的，以下是利用栈和队列判断回文的代码，供您参考：

def is_palindrome(s):
    # 创建栈和队列
    stack = []
    queue = []
    # 将字符串中的字符依次入栈和入队
    for c in s:
        if c.isalpha():
            stack.append(c.lower())
            queue.append(c.lower())
    # 依次比较栈和队列中的字符是否相等
    while stack and queue:
        if stack.pop() != queue.pop(0):
            return False
    return True

当然，上面这段代码只是实现了判断回文的功能，如果需要更全面的栈和队列知识，还需要更加深入的学习。

相关问题

c语言利用栈和队列完成回文的判断代码

好的，以下是 C 语言利用栈和队列实现回文判断的代码示例：

#include<stdio.h>
#include<stdlib.h>

#define MAXSIZE 100

typedef int ElemType;

typedef struct
{
    ElemType data[MAXSIZE];
    int top;
}SqStack;

typedef struct QNode
{
    ElemType data;
    struct QNode *next;
}QNode, *QueuePtr;

typedef struct
{
    QueuePtr front, rear;
}LinkQueue;

void InitStack(SqStack *S)
{
    S->top = -1;
}

int StackEmpty(SqStack S)
{
    if(S.top == -1) return 1;
    else return 0;
}

int StackFull(SqStack S)
{
    if(S.top == MAXSIZE - 1) return 1;
    else return 0;
}

void Push(SqStack *S, ElemType x)
{
    if(StackFull(*S))
    {
        printf("Stack is full!\n");
        exit(1);
    }
    else
    {
        S->data[++S->top] = x;
    }
}

ElemType Pop(SqStack *S)
{
    if(StackEmpty(*S))
    {
        printf("Stack is empty!\n");
        exit(1);
    }
    else
    {
        return S->data[S->top--];
    }
}

void InitQueue(LinkQueue *Q)
{
    Q->front = Q->rear = (QueuePtr)malloc(sizeof(QNode));
    if(!Q->front)
    {
        printf("Memory allocation failed!\n");
        exit(1);
    }
    Q->front->next = NULL;
}

void EnQueue(LinkQueue *Q, ElemType x)
{
    QueuePtr p;
    p = (QueuePtr)malloc(sizeof(QNode));
    if(!p)
    {
        printf("Memory allocation failed!\n");
        exit(1);
    }
    p->data = x;
    p->next = NULL;
    Q->rear->next = p;
    Q->rear = p;
}

ElemType DeQueue(LinkQueue *Q)
{
    QueuePtr p;
    ElemType x;
    if(Q->front == Q->rear)
    {
        printf("Queue is empty!\n");
        exit(1);
    }
    p = Q->front->next;
    Q->front->next = p->next;
    x = p->data;
    if(Q->rear == p)
        Q->rear = Q->front;
    free(p);
    return x;
}

int main()
{
    SqStack s;
    LinkQueue q;
    int n, i, flag = 1;
    char c[MAXSIZE];

    printf("Input a string:");
    gets(c);

    n = strlen(c);

    InitStack(&s);
    InitQueue(&q);

    for(i = 0; i < n; i++)
    {
        if(c[i] >= 'A' && c[i] <= 'Z')
            c[i] = c[i] + 32;
        if(c[i] >= 'a' && c[i] <= 'z')
        {
            Push(&s, c[i]);
            EnQueue(&q, c[i]);
        }
    }

    while(!StackEmpty(s) && flag)
    {
        if(Pop(&s) != DeQueue(&q))
            flag = 0;
    }

    if(flag == 1)
        printf("The string is a palindrome!\n");
    else
        printf("The string is not a palindrome!\n");

    return 0;
}

请问您还有什么需要我帮忙的吗？

利用栈和队列完成回文的判断代码c++

栈和队列都可以用来实现回文的判断。以下是两种方法的C++代码实现：

使用栈：

#include <iostream>
#include <stack>
#include <string>
using namespace std;

bool isPalindrome(string str) {
    stack<char> s;
    int n = str.length();
    int i;

    for (i = 0; i < n / 2; i++) {
        s.push(str[i]);
    }

    if (n % 2 != 0) {
        i++;
    }

    while (i < n) {
        if (s.top() != str[i]) {
            return false;
        }
        s.pop();
        i++;
    }

    return true;
}

int main() {
    string str;
    cout << "Enter a string: ";
    getline(cin, str);

    if (isPalindrome(str)) {
        cout << str << " is a palindrome" << endl;
    } else {
        cout << str << " is not a palindrome" << endl;
    }

    return 0;
}

使用队列：

#include <iostream>
#include <queue>
#include <string>
using namespace std;

bool isPalindrome(string str) {
    queue<char> q;
    int n = str.length();
    int i;

    for (i = n - 1; i >= n / 2; i--) {
        q.push(str[i]);
    }

    if (n % 2 != 0) {
        i--;
    }

    while (i >= 0) {
        if (q.front() != str[i]) {
            return false;
        }
        q.pop();
        i--;
    }

    return true;
}

int main() {
    string str;
    cout << "Enter a string: ";
    getline(cin, str);

    if (isPalindrome(str)) {
        cout << str << " is a palindrome" << endl;
    } else {
        cout << str << " is not a palindrome" << endl;
    }

    return 0;
}