使用堆栈检查给定的字符串是否为回文

给定字符串str ，任务是使用堆栈查找给定的字符串是否是回文。
例子：

Input: str = “geeksforgeeks” 
Output: No
Input: str = “madam” 
Output: Yes 
 

方法：

• 找到字符串len的长度。现在，找到mid = len / 2 。
• 将所有元素压入堆栈的中间，即str [0…mid-1] 。
• 如果字符串的长度是奇数，则忽略中间字符。
• 直到字符串的末尾，保留从堆栈中弹出的元素，并将其与当前字符(即字符串[i])进行比较。
• 如果不匹配，则该字符串不是回文。如果所有元素都匹配，则该字符串为回文。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true
// if string is a palindrome
bool isPalindrome(string s)
{
    int length = s.size();
 
    // Creating a Stack
    stack st;
 
    // Finding the mid
    int i, mid = length / 2;
 
    for (i = 0; i < mid; i++) {
        st.push(s[i]);
    }
 
    // Checking if the length of the string
    // is odd, if odd then neglect the
    // middle character
    if (length % 2 != 0) {
        i++;
    }
   
    char ele;
    // While not the end of the string
    while (s[i] != '\0')
    {
         ele = st.top();
         st.pop();
 
    // If the characters differ then the
    // given string is not a palindrome
    if (ele != s[i])
        return false;
        i++;
    }
 
return true;
}
 
// Driver code
int main()
{
    string s = "madam";
 
    if (isPalindrome(s)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
 
    return 0;
}
 
// This Code is Contributed by Harshit Srivastava

C

// C implementation of the approach
#include 
#include 
#include 
#include 
 
char* stack;
int top = -1;
 
// push function
void push(char ele)
{
    stack[++top] = ele;
}
 
// pop function
char pop()
{
    return stack[top--];
}
 
// Function that returns 1
// if str is a palindrome
int isPalindrome(char str[])
{
    int length = strlen(str);
 
    // Allocating the memory for the stack
    stack = (char*)malloc(length * sizeof(char));
 
    // Finding the mid
    int i, mid = length / 2;
 
    for (i = 0; i < mid; i++) {
        push(str[i]);
    }
 
    // Checking if the length of the string
    // is odd, if odd then neglect the
    // middle character
    if (length % 2 != 0) {
        i++;
    }
 
    // While not the end of the string
    while (str[i] != '\0') {
        char ele = pop();
 
        // If the characters differ then the
        // given string is not a palindrome
        if (ele != str[i])
            return 0;
        i++;
    }
 
    return 1;
}
 
// Driver code
int main()
{
    char str[] = "madam";
 
    if (isPalindrome(str)) {
        printf("Yes");
    }
    else {
        printf("No");
    }
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
static char []stack;
static int top = -1;
 
// push function
static void push(char ele)
{
    stack[++top] = ele;
}
 
// pop function
static char pop()
{
    return stack[top--];
}
 
// Function that returns 1
// if str is a palindrome
static int isPalindrome(char str[])
{
    int length = str.length;
 
    // Allocating the memory for the stack
    stack = new char[length * 4];
 
    // Finding the mid
    int i, mid = length / 2;
 
    for (i = 0; i < mid; i++)
    {
        push(str[i]);
    }
 
    // Checking if the length of the String
    // is odd, if odd then neglect the
    // middle character
    if (length % 2 != 0)
    {
        i++;
    }
 
    // While not the end of the String
    while (i < length)
    {
        char ele = pop();
 
        // If the characters differ then the
        // given String is not a palindrome
        if (ele != str[i])
            return 0;
        i++;
    }
 
    return 1;
}
 
// Driver code
public static void main(String[] args)
{
    char str[] = "madam".toCharArray();
 
    if (isPalindrome(str) == 1)
    {
        System.out.printf("Yes");
    }
    else
    {
        System.out.printf("No");
    }
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
stack = []
top = -1
 
# push function
def push(ele: str):
    global top
    top += 1
    stack[top] = ele
 
# pop function
def pop():
    global top
    ele = stack[top]
    top -= 1
    return ele
 
# Function that returns 1
# if str is a palindrome
def isPalindrome(string: str) -> bool:
    global stack
    length = len(string)
 
    # Allocating the memory for the stack
    stack = ['0'] * (length + 1)
 
    # Finding the mid
    mid = length // 2
    i = 0
    while i < mid:
        push(string[i])
        i += 1
 
    # Checking if the length of the string
    # is odd, if odd then neglect the
    # middle character
    if length % 2 != 0:
        i += 1
 
    # While not the end of the string
    while i < length:
        ele = pop()
 
        # If the characters differ then the
        # given string is not a palindrome
        if ele != string[i]:
            return False
        i += 1
    return True
 
# Driver Code
if __name__ == "__main__":
 
    string = "madam"
 
    if isPalindrome(string):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# sanjeev2552

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
static char []stack;
static int top = -1;
 
// push function
static void push(char ele)
{
    stack[++top] = ele;
}
 
// pop function
static char pop()
{
    return stack[top--];
}
 
// Function that returns 1
// if str is a palindrome
static int isPalindrome(char []str)
{
    int length = str.Length;
 
    // Allocating the memory for the stack
    stack = new char[length * 4];
 
    // Finding the mid
    int i, mid = length / 2;
 
    for (i = 0; i < mid; i++)
    {
        push(str[i]);
    }
 
    // Checking if the length of the String
    // is odd, if odd then neglect the
    // middle character
    if (length % 2 != 0)
    {
        i++;
    }
 
    // While not the end of the String
    while (i < length)
    {
        char ele = pop();
 
        // If the characters differ then the
        // given String is not a palindrome
        if (ele != str[i])
            return 0;
        i++;
    }
    return 1;
}
 
// Driver code
public static void Main(String[] args)
{
    char []str = "madam".ToCharArray();
 
    if (isPalindrome(str) == 1)
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by 29AjayKumar

输出：

Yes

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