<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 validPalindrome(string s) {
    int left = 0;
    int right = s.length() - 1;
    bool isDel = false;
    bool isRight = false;
    bool isLeft = false;
    int save_left = left;
    int save_right = right;
    while (left < right)
    {
      if (s[left] != s[right])
      {
        if (isDel)
        {
          if (isRight)
          {
            if (isLeft)
            {
              return false;
            }
            left = save_left;
            right = save_right;
          }
          else
          {
            return false;
          }
        }

        if (!isRight && s[left] == s[right - 1])
        {
          save_left = left;
          save_right = right;
          isRight = true;
          right--;
        }
        else if (s[left + 1] == s[right])
        {
          isLeft = true;
          left++;
        }
        else
        {
          return false;
        }
        isDel = true;
        continue;
      }
      left++;
      right--;
    }
    return true;
  }
};
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