<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;     //array(<vector這種)是用sizeof(),string是用length()
        while(left <= right) {
            if(s[left] != s[right]) return isPalindrome(s,left,right-1) || isPalindrome(s,left+1,right);    //如果對不上就把左邊的數刪掉再比較或把右邊的數刪掉再比較
            else {
                left++;
                right--;
            }
        }
        return true;
    }
private:
    bool isPalindrome(string s,int left,int right) {
        while(left <= right) {
            if(s[left] != s[right]) return false;
            else {
                left++;
                right--;
            }
        }
        return true;
    }
};
23624