<p>Given a string <code>s</code>, return <code>true</code> <em>if the </em><code>s</code><em> can be palindrome after deleting <strong>at most one</strong> character from it</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abca&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You could delete the character &#39;c&#39;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abc&quot;
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:

    bool isPalindrom(string& s)
    {
        for(int i = 0; i < s.size() / 2; ++i)
        {
            if(s[i] != s[s.size() - i - 1])
                return false;
        } 
        return true;
    }

    bool validPalindrome(string s) {
        string s_orig(s), cur;
        int len = s_orig.size();
        for(int i = 0; i < len / 2; ++i)
        {
            if(s_orig[i] != s_orig[len - i - 1])
            {
                s = s_orig;
                cur = s.erase(i, 1);
                if(isPalindrom(cur))
                    return true;
                s = s_orig;
                cur = s.erase(len - i - 1, 1);
                if(isPalindrom(cur))
                    return true;
                return false;
            }
        }
        return true;
    }
};
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