用c语言利用栈结构判断回文

可以使用栈结构来判断回文。具体方法是将字符串的前一半依次入栈，然后再将后一半与栈中的元素依次比较，如果全部相等，则说明该字符串是回文。以下是示例代码：

#include <stdio.h>
#include <string.h>

#define MAX_SIZE 100

int main()
{
char str[MAX_SIZE];
int len, i, mid;
char stack[MAX_SIZE];
int top = -1;

printf("请输入一个字符串：");
scanf("%s", str);

len = strlen(str);
mid = len / 2;

for (i = 0; i < mid; i++) {
stack[++top] = str[i];
}

if (len % 2 == 1) {
i = mid + 1;
} else {
i = mid;
}

while (i < len) {
if (stack[top--] != str[i++]) {
printf("%s 不是回文\n", str);
return 0;
}
}

printf("%s 是回文\n", str);

return 0;
}

相关问题

用c语言输入一段数据并利用栈结构判断回文

可以使用以下代码实现：

#include <stdio.h>
#include <string.h>

#define MAX_SIZE 100

int top = -1;
char stack[MAX_SIZE];

void push(char c) {
if (top == MAX_SIZE - 1) {
printf("Stack Overflow\n");
return;
}
stack[++top] = c;
}

char pop() {
if (top == -1) {
printf("Stack Underflow\n");
return '\0';
}
return stack[top--];
}

int main() {
char str[MAX_SIZE];
int i, len, mid;

printf("Enter a string: ");
scanf("%s", str);

len = strlen(str);
mid = len / 2;

for (i = 0; i < mid; i++) {
push(str[i]);
}

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

while (str[i] != '\0') {
if (pop() != str[i]) {
printf("%s is not a palindrome\n", str);
return 0;
}
i++;
}

printf("%s is a palindrome\n", str);

return 0;
}

C语言数据结构和ACM2023-第四章-循环队列判断回文- LJD 【问题描述】 回文判断：假设称正读和反读都相同的字符序列为“回文”，例如‘abba’和‘abcba’是回文，‘abcde’和‘ababab’则不是。试写一个算法判别读入的一个以‘@’结束符的字符序列是否是“回文”。 提示：利用栈和队列知识。 Copying judgment: Assuming that the character sequence known as the same reading and anti-reading is "replying", such as 'abba' and 'abcba' are repayment, 'Abcde' and 'Ababab' are not. Try to write a character sequence of an algorithm to read in the '@' ending character. Tip: Use stacks and queue knowledge. 【输入形式】 以@结尾的字符串 【输出形式】 判断结果 【样例输入】 abcdcba@ 【样例输出】 此字符串为回文

以下是C语言的实现，使用循环队列和栈结构判断回文：

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

#define MAX_SIZE 100

typedef struct {
    char data[MAX_SIZE];
    int front, rear;
} Queue;

typedef struct {
    char data[MAX_SIZE];
    int top;
} Stack;

void initQueue(Queue *q) {
    q->front = q->rear = 0;
}

void initStack(Stack *s) {
    s->top = -1;
}

int isEmptyQueue(Queue *q) {
    return q->front == q->rear;
}

int isFullQueue(Queue *q) {
    return q->front == (q->rear + 1) % MAX_SIZE;
}

int isEmptyStack(Stack *s) {
    return s->top == -1;
}

int isFullStack(Stack *s) {
    return s->top == MAX_SIZE - 1;
}

void enqueue(Queue *q, char c) {
    if (isFullQueue(q)) {
        printf("Queue is full\n");
        exit(1);
    }
    q->data[q->rear] = c;
    q->rear = (q->rear + 1) % MAX_SIZE;
}

char dequeue(Queue *q) {
    if (isEmptyQueue(q)) {
        printf("Queue is empty\n");
        exit(1);
    }
    char c = q->data[q->front];
    q->front = (q->front + 1) % MAX_SIZE;
    return c;
}

void push(Stack *s, char c) {
    if (isFullStack(s)) {
        printf("Stack is full\n");
        exit(1);
    }
    s->data[++s->top] = c;
}

char pop(Stack *s) {
    if (isEmptyStack(s)) {
        printf("Stack is empty\n");
        exit(1);
    }
    return s->data[s->top--];
}

int main() {
    char str[MAX_SIZE];
    printf("Input a string ending with '@': ");
    scanf("%s", str);

    int length = strlen(str);
    if (str[length - 1] != '@') {
        printf("Invalid input\n");
        return 0;
    }

    Queue q;
    Stack s;
    initQueue(&q);
    initStack(&s);

    int i;
    for (i = 0; i < length - 1; i++) {
        enqueue(&q, str[i]);
        push(&s, str[i]);
    }

    while (!isEmptyQueue(&q) && !isEmptyStack(&s)) {
        if (dequeue(&q) != pop(&s)) {
            printf("This string is not a palindrome\n");
            return 0;
        }
    }

    printf("This string is a palindrome\n");

    return 0;
}

以上程序可以根据输入的字符串进行回文判断，如果是回文则输出"This string is a palindrome"，否则输出"This string is not a palindrome"。