检查字符串中的字符是否在 O(1) 额外空间中形成回文

给定字符串str 。字符串可能包含小写字母、特殊字符、数字，甚至空格。任务是检查是否只有字符串中存在的字母形成回文组合，而不使用任何额外的空间。
注意：不允许使用额外的空间来解决这个问题。此外，字符串中的字母是小写的，并且字符串可能包含特殊字符、数字，甚至空格以及小写字母。

例子：

Input : str = “m a 343 la y a l am” 
Output : YES 
The characters in the string form the sequence “malayalam”, which is a palindrome.

Input : str = “malayalam” 
Output : YES

方法：

• 创建两个实用函数来获取字符串中字符的第一个和最后一个位置。
• 开始遍历字符串，每次都不断寻找第一个和最后一个字符的位置。
• 如果每次迭代的第一个和最后一个字符都相同并且字符串被完全遍历，则打印YES，否则打印NO。

下面是上述方法的实现：

C++

// CPP program to check if the characters in
// the given string forms a Palindrome in
// O(1) extra space
#include 
using namespace std;
 
// Utility function to get the position of
// first character in the string
int firstPos(string str, int start, int end)
{
    int firstChar = -1;
 
    // Get the position of first character
    // in the string
    for (int i = start; i <= end; i++) {
        if (str[i] >= 'a' && str[i] <= 'z') {
            firstChar = i;
            break;
        }
    }
 
    return firstChar;
}
 
// Utility function to get the position of
// last character in the string
int lastPos(string str, int start, int end)
{
    int lastChar = -1;
 
    // Get the position of last character
    // in the string
    for (int i = start; i >= end; i--) {
        if (str[i] >= 'a' && str[i] <= 'z') {
            lastChar = i;
            break;
        }
    }
 
    return lastChar;
}
 
// Function to check if the characters in
// the given string forms a Palindrome in
// O(1) extra space
bool isPalindrome(string str)
{
    int firstChar = 0, lastChar = str.length() - 1;
    bool ch = true;
 
    for (int i = 0; i < str.length(); i++) {
        firstChar = firstPos(str, firstChar, lastChar);
        lastChar = lastPos(str, lastChar, firstChar);
 
        // break, when all letters are checked
        if (lastChar < 0 || firstChar < 0)
            break;
        if (str[firstChar] == str[lastChar]) {
            firstChar++;
            lastChar--;
            continue;
        }
 
        // if mismatch found, break the loop
        ch = false;
        break;
    }
 
    return (ch);
}
 
// Driver code
int main()
{
    string str = "m     a  343 la y a l am";
    if (isPalindrome(str))
        cout << "YES";
    else
        cout << "NO";
 
    return 0;
}

Java

// Java program to check if
// the characters in the given
// string forms a Palindrome
// in O(1) extra space
import java.io.*;
 
class GFG
{
 
// Utility function to get
// the position of first
// character in the string
static int firstPos(String str,
                    int start,
                    int end)
{
    int firstChar = -1;
     
    // Get the position of
    // first character in
    // the string
    for(int i = start; i <= end; i++)
    {
        if(str.charAt(i) >= 'a'&&
        str.charAt(i) <= 'z')
        {
            firstChar = i;
            break;
        }
    }
     
    return firstChar;
}
 
// Utility function to get
// the position of last
// character in the string
static int lastPos(String str,
                int start,
                int end)
{
    int lastChar = -1;
     
    // Get the position of last
    // character in the string
    for(int i = start; i >= end; i--)
    {
        if(str.charAt(i) >= 'a'&&
        str.charAt(i) <= 'z')
        {
            lastChar = i;
            break;
        }
    }
     
    return lastChar;
}
 
// Function to check if the
// characters in the given
// string forms a Palindrome
// in O(1) extra space
static boolean isPalindrome(String str)
{
    int firstChar = 0,
        lastChar = str.length() - 1;
    boolean ch = true;
 
    for (int i = 0; i < str.length(); i++)
    {
        firstChar = firstPos(str, firstChar,
                                lastChar);
        lastChar = lastPos(str, lastChar,
                                firstChar);
 
        // break, when all
        // letters are checked
        if (lastChar < 0 || firstChar < 0)
            break;
        if (str.charAt(firstChar) ==
            str.charAt(lastChar))
        {
            firstChar++;
            lastChar--;
            continue;
        }
         
        // if mismatch found,
        // break the loop
        ch = false;
        break;
    }
 
        return ch;
 
}
 
// Driver code
public static void main (String[] args)
{
    String str = "m a 343 la y a l am";
     
    if(isPalindrome(str))
        System.out.print("YES");
    else
    System.out.println("NO");
}
}
 
// This code is contributed
// by inder_verma.

Python 3

# Python 3 program to check if the characters
# in the given string forms a Palindrome in
# O(1) extra space
 
# Utility function to get the position of
# first character in the string
def firstPos(str, start, end):
 
    firstChar = -1
 
    # Get the position of first character
    # in the string
    for i in range(start, end + 1):
        if (str[i] >= 'a' and str[i] <= 'z') :
            firstChar = i
            break
 
    return firstChar
 
# Utility function to get the position of
# last character in the string
def lastPos(str, start, end):
 
    lastChar = -1
 
    # Get the position of last character
    # in the string
    for i in range(start, end - 1, -1) :
        if (str[i] >= 'a' and str[i] <= 'z') :
            lastChar = i
            break
 
    return lastChar
 
# Function to check if the characters in
# the given string forms a Palindrome in
# O(1) extra space
def isPalindrome(str):
 
    firstChar = 0
    lastChar = len(str) - 1
    ch = True
 
    for i in range(len(str)) :
        firstChar = firstPos(str, firstChar, lastChar);
        lastChar = lastPos(str, lastChar, firstChar);
 
        # break, when all letters are checked
        if (lastChar < 0 or firstChar < 0):
            break
        if (str[firstChar] == str[lastChar]):
            firstChar += 1
            lastChar -= 1
            continue
 
        # if mismatch found, break the loop
        ch = False
        break
 
    return (ch)
 
# Driver code
if __name__ == "__main__":
     
    str = "m     a 343 la y a l am"
    if (isPalindrome(str)):
        print("YES")
    else:
        print("NO")
 
# This code is contributed by ita_c

C#

// C# program to check if
// the characters in the given
// string forms a Palindrome
// in O(1) extra space
using System;
 
class GFG
{
 
// Utility function to get
// the position of first
// character in the string
static int firstPos(string str,
                    int start,
                    int end)
{
    int firstChar = -1;
     
    // Get the position of
    // first character in
    // the string
    for(int i = start; i <= end; i++)
    {
        if(str[i] >= 'a'&&
           str[i] <= 'z')
        {
            firstChar = i;
            break;
        }
    }
     
    return firstChar;
}
 
// Utility function to get
// the position of last
// character in the string
static int lastPos(string str,
                   int start,
                   int end)
{
    int lastChar = -1;
     
    // Get the position of last
    // character in the string
    for(int i = start; i >= end; i--)
    {
        if(str[i] >= 'a'&&
           str[i] <= 'z')
        {
            lastChar = i;
            break;
        }
    }
     
    return lastChar;
}
 
// Function to check if the
// characters in the given
// string forms a Palindrome
// in O(1) extra space
static bool isPalindrome(string str)
{
    int firstChar = 0,
        lastChar = str.Length - 1;
    bool ch = true;
 
    for (int i = 0; i < str.Length; i++)
    {
        firstChar = firstPos(str, firstChar,
                                  lastChar);
        lastChar = lastPos(str, lastChar,
                                firstChar);
 
        // break, when all
        // letters are checked
        if (lastChar < 0 || firstChar < 0)
            break;
        if (str[firstChar] ==
            str[lastChar])
        {
            firstChar++;
            lastChar--;
            continue;
        }
         
        // if mismatch found,
        // break the loop
        ch = false;
        break;
    }
 
        return ch;
}
 
// Driver code
public static void Main ()
{
    string str = "m a 343 la y a l am";
     
    if(isPalindrome(str))
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}
 
// This code is contributed
// by inder_verma.

PHP

= 'a' and
            $str[$i] <= 'z')
        {
            $firstChar = $i;
            break;
        }
    }
 
    return $firstChar;
}
 
// Utility function to get the
// position of last character
// in the string
function lastPos($str, $start, $end)
{
    $lastChar = -1;
 
    // Get the position of last
    // character in the string
    for ( $i = $start; $i >= $end; $i--)
    {
        if ($str[$i] >= 'a' and
            $str[$i] <= 'z')
        {
            $lastChar = $i;
            break;
        }
    }
 
    return $lastChar;
}
 
// Function to check if the
// characters in the given
// string forms a Palindrome
// in O(1) extra space
function isPalindrome($str)
{
    $firstChar = 0;
    $lastChar = count($str) - 1;
    $ch = true;
 
    for ($i = 0; $i < count($str); $i++)
    {
        $firstChar = firstPos($str, $firstChar,
                                    $lastChar);
        $lastChar = lastPos($str, $lastChar,
                                  $firstChar);
 
        // break, when all letters are checked
        if ($lastChar < 0 or $firstChar < 0)
            break;
        if ($str[$firstChar] == $str[$lastChar])
        {
            $firstChar++;
            $lastChar--;
            continue;
        }
 
        // if mismatch found,
        // break the loop
        $ch = false;
        break;
    }
 
    return ($ch);
}
 
// Driver code
$str = "m a 343 la y a l am";
if (isPalindrome($str))
    echo "YES";
else
    echo "NO";
 
// This code is contributed
// by inder_verma.
?>

Javascript

// Javascript program to check if
// the characters in the given
// string forms a Palindrome
// in O(1) extra space
 
// Utility function to get
// the position of first
// character in the string
function firstPos(str,start,end)
{
    let firstChar = -1;
       
    // Get the position of
    // first character in
    // the string
    for(let i = start; i <= end; i++)
    {
        if(str[i] >= 'a'&&
        str[i] <= 'z')
        {
            firstChar = i;
            break;
        }
    }
       
    return firstChar;
}
 
// Utility function to get
// the position of last
// character in the string
function lastPos(str,start,end)
{
    let lastChar = -1;
       
    // Get the position of last
    // character in the string
    for(let i = start; i >= end; i--)
    {
        if(str[i] >= 'a'&&
        str[i] <= 'z')
        {
            lastChar = i;
            break;
        }
    }
       
    return lastChar;
}
 
   
// Function to check if the
// characters in the given
// string forms a Palindrome
// in O(1) extra space
function isPalindrome(str)
{
    let firstChar = 0,
        lastChar = str.length - 1;
    let ch = true;
   
    for (let i = 0; i < str.length; i++)
    {
        firstChar = firstPos(str, firstChar,
                                lastChar);
        lastChar = lastPos(str, lastChar,
                                firstChar);
   
        // break, when all
        // letters are checked
        if (lastChar < 0 || firstChar < 0)
            break;
        if (str[firstChar] ==
            str[lastChar])
        {
            firstChar++;
            lastChar--;
            continue;
        }
           
        // if mismatch found,
        // break the loop
        ch = false;
        break;
    }
   
        return ch;
}
 
// Driver code
let str = "m a 343 la y a l am";
if(isPalindrome(str))
    document.write("YES");
else
    document.write("NO");
 
     
 
// This code is contributed by avanitrachhadiya2155

输出：

YES