给定一个字符串str ，任务是从字符串str 中删除所有连续的重复项，并检查最终字符串是否为回文。如果是回文则打印“Yes” ，否则打印“No” 。
例子：
Input: str = “abbcbbbaaa”
Output: Yes
Explanation:
On removing all consecutive duplicates characters, the string becomes “abcba” which is a palindrome.
Input: str = “aaabbbaaccc”
Output: No
Explanation:
On removing all consecutive duplicates characters, the string becomes “abac” which is not a palindrome.
方法：我们的想法是创建一个从给定的字符串，并检查了新的字符串，如果新的字符串是回文与否。以下是步骤：
- 初始化新字符串newStr = “” 。
- 逐个遍历给定字符串的所有字符，如果当前字符与前一个字符不同，则将此字符附加到新字符串newStr 。
- 否则检查下一个字符。
- 检查最终形成的字符串是否为回文。如果是回文则打印“Yes” ，否则打印“No” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if a string
// is palindrome or not
bool isPalindrome(string str)
{
// Length of the string
int len = str.length();
// Check if its a palindrome
for (int i = 0; i < len; i++) {
// If the palindromic
// condition is not met
if (str[i] != str[len - i - 1])
return false;
}
// Return true as str is palindromic
return true;
}
// Function to check if string str is
// palindromic after removing every
// consecutive characters from the str
bool isCompressablePalindrome(string str)
{
// Length of the string str
int len = str.length();
// Create an empty
// compressed string
string compressed = "";
// The first character will
// always be included in
// the final string
compressed.push_back(str[0]);
// Check all the characters
// of the string
for (int i = 1; i < len; i++) {
// If the current character
// is not same as its previous
// one, then insert it in the
// final string
if (str[i] != str[i - 1])
compressed.push_back(str[i]);
}
// Check if the compressed
// string is a palindrome
return isPalindrome(compressed);
}
// Driver Code
int main()
{
// Given string
string str = "abbcbbbaaa";
// Function call
if (isCompressablePalindrome(str))
cout << "Yes\n";
else
cout << "No\n";
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if a String
// is palindrome or not
static boolean isPalindrome(String str)
{
// Length of the String
int len = str.length();
// Check if its a palindrome
for (int i = 0; i < len; i++)
{
// If the palindromic
// condition is not met
if (str.charAt(i) != str.charAt(len - i - 1))
return false;
}
// Return true as str is palindromic
return true;
}
// Function to check if String str is
// palindromic after removing every
// consecutive characters from the str
static boolean isCompressablePalindrome(String str)
{
// Length of the String str
int len = str.length();
// Create an empty
// compressed String
String compressed = "";
// The first character will
// always be included in
// the final String
compressed = String.valueOf(str.charAt(0));
// Check all the characters
// of the String
for (int i = 1; i < len; i++)
{
// If the current character
// is not same as its previous
// one, then insert it in the
// final String
if (str.charAt(i) != str.charAt(i - 1))
compressed += str.charAt(i);
}
// Check if the compressed
// String is a palindrome
return isPalindrome(compressed);
}
// Driver Code
public static void main(String[] args)
{
// Given String
String str = "abbcbbbaaa";
// Function call
if (isCompressablePalindrome(str))
System.out.print("Yes\n");
else
System.out.print("No\n");
}
}
// This code is contributed by amal kumar choubeyPython3
# Python3 program for the above approach
# Function to check if a string
# is palindrome or not
def isPalindrome(Str):
# Length of the string
Len = len(Str)
# Check if its a palindrome
for i in range(Len):
# If the palindromic
# condition is not met
if (Str[i] != Str[Len - i - 1]):
return False
# Return true as Str is palindromic
return True
# Function to check if string str is
# palindromic after removing every
# consecutive characters from the str
def isCompressablePalindrome(Str):
# Length of the string str
Len = len(Str)
# Create an empty compressed string
compressed = ""
# The first character will always
# be included in final string
compressed += Str[0]
# Check all the characters
# of the string
for i in range(1, Len):
# If the current character
# is not same as its previous
# one, then insert it in
# the final string
if (Str[i] != Str[i - 1]):
compressed += Str[i]
# Check if the compressed
# string is a palindrome
return isPalindrome(compressed)
# Driver Code
if __name__ == '__main__':
# Given string
Str = "abbcbbbaaa"
# Function call
if (isCompressablePalindrome(Str)):
print("Yes")
else:
print("No")
# This code is contributed by himanshu77C#
// C# program for the above approach
using System;
class GFG{
// Function to check if a String
// is palindrome or not
static bool isPalindrome(String str)
{
// Length of the String
int len = str.Length;
// Check if its a palindrome
for(int i = 0; i < len; i++)
{
// If the palindromic
// condition is not met
if (str[i] != str[len - i - 1])
return false;
}
// Return true as str is palindromic
return true;
}
// Function to check if String str is
// palindromic after removing every
// consecutive characters from the str
static bool isCompressablePalindrome(String str)
{
// Length of the String str
int len = str.Length;
// Create an empty
// compressed String
String compressed = "";
// The first character will
// always be included in
// the readonly String
compressed = String.Join("", str[0]);
// Check all the characters
// of the String
for(int i = 1; i < len; i++)
{
// If the current character
// is not same as its previous
// one, then insert it in the
// readonly String
if (str[i] != str[i - 1])
compressed += str[i];
}
// Check if the compressed
// String is a palindrome
return isPalindrome(compressed);
}
// Driver Code
public static void Main(String[] args)
{
// Given String
String str = "abbcbbbaaa";
// Function call
if (isCompressablePalindrome(str))
Console.Write("Yes\n");
else
Console.Write("No\n");
}
}
// This code is contributed by amal kumar choubey输出：
Yes
时间复杂度： O(N) ，其中 N 是字符串的长度。
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