判断回文字符串

LRH1993 · LRH1993 · commit 59857909a1d7 · 2017-10-13T10:07:32.000+08:00
diff --git a/SUMMARY.md b/SUMMARY.md
@@ -201,6 +201,7 @@
   - [String](/algorithm/LeetCode/String.md)
     - [Restore IP Addresses](/algorithm/LeetCode/String/ip.md)
     - [Rotate String](/algorithm/LeetCode/String/Rotate-String.md)
+    - [Valid Palindrome](/algorithm/LeetCode/String/Valid-Palindrome.md)
 
 ## 设计模式
 
diff --git a/algorithm/LeetCode/String/Valid-Palindrome.md b/algorithm/LeetCode/String/Valid-Palindrome.md
@@ -0,0 +1,57 @@
+## 一、题目
+
+> Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
+>
+> For example,
+> `"A man, a plan, a canal: Panama"` is a palindrome.
+> `"race a car"` is *not* a palindrome.
+>
+> **Note:**
+>
+> Have you consider that the string might be empty? This is a good question to ask during an interview.
+>
+> For the purpose of this problem, we define empty string as valid palindrome.
+
+判断一个字符串是不是回文串。
+
+## 二、解题思路
+
+字符串的回文判断问题，由于字符串可随机访问，故逐个比较首尾字符是否相等最为便利，即常见的『两根指针』技法。
+
+两步走：
+
+1. 找到最左边和最右边的第一个合法字符(字母或者字符)
+2. 一致转换为小写进行比较
+
+## 三、解题代码
+
+```java
+public class Solution {
+    public boolean isPalindrome(String s) {
+        if (s == null || s.trim().isEmpty()) {
+            return true;
+        }
+
+        int l = 0, r = s.length() - 1;
+        while (l < r) {
+            if(!Character.isLetterOrDigit(s.charAt(l))) {
+                l++;
+                continue;
+            }
+            if(!Character.isLetterOrDigit(s.charAt(r))) {
+                r--;
+                continue;
+            }
+            if (Character.toLowerCase(s.charAt(l)) == Character.toLowerCase(s.charAt(r))) {
+                l++;
+                r--;
+            } else {
+                return false;
+            }
+        }
+
+        return true;
+    }
+}
+```
+
