Palindrome Index | Lei Jiang Coding

Palindrome Index | Lei Jiang Coding
You are given a string of lower case letters. Your task is to figure out the index of the character on whose removal it will make the string a palindrome.
There will always be a valid solution.
In case the string is already a palindrome, then -1 is also a valid answer along with possible indices.
http://www.cnblogs.com/lautsie/p/3911501.html

        string s;

        cin >> s;

        int l = 0;

        int r = s.size() - 1;

        while (l < r && s[l] == s[r]) {

            l++;

            r--;

        }

        if (l >= r) {

            cout << -1 << endl;

            continue;

        }

        int ll = l + 1;

        int rr = r;

        while (ll < rr && s[ll] == s[rr]) {

            ll++;

            rr--;

        }

        if (ll >= rr) {

            cout << l << endl;

        } else {

            cout << r << endl;

        }

int palindromeIndex( string s ){

    int start = 0, end = s.size() - 1;

    bool flag = false;

    while (start < end && s[ start ] == s[ end ]){

        start++;

        end--;

    }

    if ( start >= end ) return -1;//palindrome

     

    if  (s [ start + 1 ] != s [ end ])

        return end;

    else if ( s [ start ] != s [ end - 1] )

        return start;

    else{

        int begin = start, last = end, left = start + 1, right = end - 1;

        while(s[ begin ] == s[ right ] && s[ left ] == s[ last ] ){

            begin++;

            last--;

            left++;

            right--;

        }

        return s[begin] != s[right] ? start : end;

    }

     

}

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