Merge branch 'master' of github.com:fifa0329/leetcode

Author: lingxiang.wang

PalindromeNumber.java

@@ -14,21 +14,113 @@
  */
 
 public class PalindromeNumber {
+    
+    
+    
     public boolean isPalindrome(int x) {
-        if (x < 0)
-            return false;
-        int k = 1;
-        while (x / k >= 10) {
-            k *= 10;
-        }
-        while (x >= 10) {
-            int left = x / k;
-            int right = x - x / 10 * 10;
-            if (left != right)
-                return false;
-            x = (x - x / k * k) / 10;
-            k /= 100;
-        }
-        return true;
+        //压根不会做，看了答案
+
+//下面方法是错的！
+//        if (x < 0)
+//            return false;
+//        int k = 1;
+//        while (x / k >= 10) {
+//            k *= 10;
+//        }
+//        while (x >= 10) {
+//            int left = x / k;
+//            int right = x - x / 10 * 10;
+//            if (left != right)
+//                return false;
+//            x = (x - x / k * k) / 10;
+//            k /= 100;
+//        }
+//        return true;
+        //我复制了这个答案 ，发现这个就是错的！
+        //10进制的重要的点，因为数字未知，一开始应该用while来判断出位数，通过不断的十进制乘上去
+        //第一位和最后一位的判断相等，思路还是很清晰的
+        //但是继续进行，判断是否回文，就有点乱了
+        
+        
+        
+        
+//        ------------------------
+
+        
+            //如何通过进制，来进行int的reverse
+
+/*
+
+        very nice try!
+
+                commented Oct 10, 2014 by dingou666
+
+        consider int overflow ? such as 1000000003 and 3000000001 is overflow......
+
+        commented Oct 12, 2014 by yomin
+
+        Amazing! Just add check condition as follows and u are done. if(x < 0 || x == Integer.MAX_VALUE) return false;
+
+        commented Oct 12, 2014 by erenalgan
+
+        Good point! Actually I did not consider overflow when writing the code... But it turns out that overflow is automatically handled: palindromeX will result in a negative number because of overflow, and that makes it never equals to the input number x...
+
+        commented Oct 12, 2014 by hln9319
+
+        Same concern, it will be overflow for some case.
+
+        commented Dec 2, 2014 by windkiosk
+
+        use long long to store palindromeX
+
+        commented Dec 17, 2014 by zhanglg921
+
+        OJ considers negative numbers as non-palindrome. Though, your solution does not handles this case explicitly - seems to me that it is taken care of as a side effect of your algorithm. For ex: if x = -121, palindromeX = 121 by the end of the for loop which is not equal to x (-121). But, there is really no need to go through the for loop, you can return false if x < 0. Otherwise, very neat solution!
+
+                commented Dec 29, 2014 by rainhacker
+        
+        */
+        
+        
+        
+        //Reverse Integer与这道题异曲同工
+        
+            int palindromeX = 0;
+            int inputX = x;
+            while(x>0){
+                palindromeX = palindromeX*10 + (x % 10);
+                x = x/10;
+            }
+            return palindromeX==inputX;
+        
+        
+        //要理解，纯粹的reverse integer会导致 overflow
+        //比如consider int overflow ? such as 1000000003 and 3000000001 is overflow......
+        
+        //为什么一次通过？因为测试的case不包含这种特殊例子
+        
+        
+        
+        
+        /* 一般用更大的类型去处理是一个思路！！！
+        bool isPalindrome(int x) {
+    long reverse = 0;
+    long num = abs(x);
+    while(x != 0){
+        reverse *= 10;
+        reverse += x % 10;
+        x /= 10;
+    }
+    return reverse == num;
+}
+         */
+        
+        
+        
+        
+        
+        
+        
+
     }
 }

ReverseInteger.java

@@ -19,6 +19,34 @@
 
 public class ReverseInteger {
     public int reverse(int x) {
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
+        
         int num = Math.abs(x);
         int ret = 0;
         while (num != 0) {