feat: add 009

Author: Blankj

README.md

@@ -9,6 +9,7 @@
 |1|[Two Sum][001]|Array, Hash Table|
 |7|[Reverse Integer][007]|Math|
 |8|[String to Integer (atoi)][008]|Math, String|
+|9|[Palindrome Number][009]|Math|
 
 
 ## Medium
@@ -32,3 +33,4 @@
 [001]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/001/README.md
 [007]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/007/README.md
 [008]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/008/README.md
+[009]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/009/README.md

note/009/README.md

@@ -0,0 +1,55 @@
+# [Palindrome Number](https://leetcode.com/problems/palindrome-number/)
+
+## Description
+
+Determine whether an integer is a palindrome. Do this without extra space.
+
+**spoilers:**
+
+**Some hints:**
+
+Could negative integers be palindromes? (ie, -1)
+
+If you are thinking of converting the integer to string, note the restriction of using extra space.
+
+You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?
+
+There is a more generic way of solving this problem.
+
+**Tags：** Math
+
+
+## 思路0
+
+题意是判断一个有符号整型数是否是回文，也就是逆序过来的整数和原整数相同，首先负数肯定不是，接下来我们分析一下最普通的解法，就是直接算出他的回文数，然后和给定值比较即可。
+
+``` java
+public class Solution {
+    public boolean isPalindrome(int x) {
+        if (x < 0) return false;
+        int copyX = x, reverse = 0;
+        while (copyX > 0) {
+            reverse = reverse * 10 + copyX % 10;
+            copyX /= 10;
+        }
+        return x == reverse;
+    }
+}
+```
+
+## 思路1
+
+好好思考下是否需要计算整个长度，比如1234321，其实不然，我们只需要计算一半的长度即可，就是在计算过程中的那个逆序数比不断除10的数大就结束计算即可，但是这也带来了另一个问题，比如10的倍数的数10010，它也会返回`true`，所以我们需要对10的倍数的数再加个判断即可，代码如下所示。
+
+``` java
+public boolean isPalindrome(int x) {
+    if (x < 0 || (x != 0 && x % 10 == 0)) return false;
+    int halfReverseX = 0;
+    while (x > halfReverseX) {
+        halfReverseX = halfReverseX * 10 + x % 10;
+        x /= 10;
+    }
+    return halfReverseX == x || halfReverseX / 10 == x;
+}
+
+```

project/LeetCode/leetcode/src/main/java/com/blankj/easy/_009/Solution.java

@@ -0,0 +1,42 @@
+package com.blankj.easy._009;
+
+/**
+ * <pre>
+ *     author: Blankj
+ *     blog  : http://blankj.com
+ *     time  : 2017/04/22
+ *     desc  :
+ * </pre>
+ */
+public class Solution {
+//    public boolean isPalindrome(int x) {
+//        if (x < 0) return false;
+//        int copyX = x, reverse = 0;
+//        while (copyX > 0) {
+//            reverse = reverse * 10 + copyX % 10;
+//            copyX /= 10;
+//        }
+//        return x == reverse;
+//    }
+
+    public boolean isPalindrome(int x) {
+        if (x < 0 || (x != 0 && x % 10 == 0)) return false;
+        int halfReverseX = 0;
+        while (x > halfReverseX) {
+            halfReverseX = halfReverseX * 10 + x % 10;
+            x /= 10;
+        }
+        return halfReverseX == x || halfReverseX / 10 == x;
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.isPalindrome(-1));
+        System.out.println(solution.isPalindrome(10010));
+
+        System.out.println(solution.isPalindrome(0));
+        System.out.println(solution.isPalindrome(11));
+        System.out.println(solution.isPalindrome(111));
+        System.out.println(solution.isPalindrome(222222222));
+    }
+}