ABC＃237 C

C

Original link
The main idea of the topic ： Given a string str, Determine whether you can add ’a’, Make the string become palindrome string .

The basic idea ： take str Head and tail ’a’ Remove , See if the remaining strings are palindromes . If the remaining substring is not a palindrome string , Then the conclusion is No .
Here's the thing to watch out for ： Because you can only add ’a’, So if str The head of the ’a’ The number of is larger than that of the tail ’a’ There are many of them , Then the conclusion is whether .

skill ： The judgment of palindrome string

bool isPalindrome(string str)
{
    
    string str2 = str;
    reverse(str.begin(), str.end());

    if (str == str2) return true;
    else return false;
}

Code 1： Double pointer

#include <iostream>
#include <cstring>
using namespace std;

const int N = 1000010;
char s[N];

int main()
{
    
    cin >> s;
    bool flag = 1;
    int i = 0, j = strlen(s) - 1, cnt1 = 0, cnt2 = 0;
    while (s[i] == 'a') i ++, cnt1 ++;
    while (s[j] == 'a') j --, cnt2 ++;
    
    if (cnt1 > cnt2) flag = 0;
    
    while (i < j)
    {
    
        if (i >= j) break;
        else if (s[i] != s[j]) 
        {
    
            flag = 0;
            break;
        }
        else i ++, j --;
    }
    
    if (flag) cout << "Yes";
    else cout << "No";
    return 0;
}

Code 2：string

#include <iostream>
#include <algorithm>
using namespace std;

#define all(x) (x).begin(), (x).end()

int main()
{
    
    string str, str2;
    int cnt1 = 0, cnt2 = 0;
    cin >> str;
    while (str.size() && str.back() == 'a') str.pop_back(), cnt1 ++; //  Remove the tail ‘a’
    reverse(all(str));
    while (str.size() && str.back() == 'a') str.pop_back(), cnt2 ++; // Take off the head ‘a’

    str2 = str;
    reverse(all(str));

    if (str == str2 && cnt2 <= cnt1) cout << "Yes";
    else cout << "No";
}