Create 680._Valid_Palindrome_II.md

Author: 0xkookoo

docs/Leetcode_Solutions/680._Valid_Palindrome_II.md

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+# 680. Valid Palindrome II
+
+**<font color=red>难度: Easy</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/valid-palindrome-ii/description/
+
+> 内容描述
+
+```
+
+Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.
+
+Example 1:
+Input: "aba"
+Output: True
+Example 2:
+Input: "abca"
+Output: True
+Explanation: You could delete the character 'c'.
+Note:
+The string will only contain lowercase characters a-z. The maximum length of the string is 50000.
+```
+
+## 解题方案
+
+> 思路 1
+
+想直接来个 for loop，看看对应除去该index元素之后的字符串是否为 palindrome 即可
+
+但是直接  Time Limit Exceeded 
+
+```python
+class Solution(object):
+    def validPalindrome(self, s):
+        """
+        :type s: str
+        :rtype: bool
+        """
+        for i in range(len(s)):
+            if s[:i] + s[i+1:] == s[i+1:][::-1] + s[:i][::-1]:
+                return True
+        return False
+```
+
+
+> 思路 2
+
+我们先定义一个reverse变量作为字符串s的翻转版本，例如
+
+```
+s =       'abbbbbca'
+reverse = 'acbbbbba'
+```
+
+然后我们从第一个字符开始比较，直到两边的字符不相等的时候，比如上面的例子我们就是index为1的时候不相等，所以i = 1，此时我们就会面临两个选择：
+
+1. 我们可以舍弃s中index为i的这个元素看看是否可以使其成为palindrome，即让s变成'abbbbca'，然后我们可以通过s[i+1:n-i] == reverse[i:n-i-1]来判断
+2. 我们可以舍弃reverse中index为i的这个元素（即s中index为n-1-i的这个元素），即让s变成'abbbbba'，我们可以通过s[i:n-1-i] == reverse[i+1:n-i]来判断
+
+```python
+class Solution(object):
+    def validPalindrome(self, s):
+        """
+        :type s: str
+        :rtype: bool
+        """
+        n = len(s)
+        if n < 3:
+            return True
+
+        reverse = s[::-1]
+        i = 0
+        while i < len(s) and s[i] == reverse[i]:
+            i += 1
+        return s[i:n-1-i] == reverse[i+1:n-i] or s[i+1:n-i] == reverse[i:n-i-1]
+```
+
+
+
+> 思路 3
+
+或者我们不搞reverse，直接在s上面原地判断即可
+
+
+```python
+class Solution(object):
+    def validPalindrome(self, s):
+        """
+        :type s: str
+        :rtype: bool
+        """
+        n = len(s)
+        if n < 3:
+            return True
+
+        l, r = 0, n - 1
+        while l < r and s[l] == s[r]:
+            l += 1
+            r -= 1
+        if l >= r:
+            return True
+        else:
+            return s[l+1:r+1] == s[l+1:r+1][::-1] or s[l:r] == s[l:r][::-1]
+```
+
+
+