Merge pull request apachecn#336 from jiangzhonglian/master

更新 9. Palindrome Number 回文数
594Surefour · May 20, 2020 · 4ce66dd · 4ce66dd
2 parents 671175a + 598a6fc
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diff --git a/.gitmodules b/.gitmodules
diff --git a/Middleware b/Middleware
diff --git a/docs/Algorithm/Leetcode/Python/009._Palindrome_Number.md b/docs/Algorithm/Leetcode/Python/009._Palindrome_Number.md
@@ -1,10 +1,16 @@
 # 9. Palindrome Number 回文数
 
+**<font color=green>难度: Easy</font>**
+
 ## 刷题内容
 
+> 原题连接
+
 * https://leetcode.com/problems/palindrome-number
 * https://leetcode-cn.com/problems/palindrome-number
 
+> 内容描述
+
 ```
 判断一个整数是否是回文数。回文数是指正序（从左向右）和倒序（从右向左）读都是一样的整数。
 
@@ -27,48 +33,70 @@
 你能不将整数转为字符串来解决这个问题吗？
 ```
 
-## 难度：Medium
-
-> 思路1 （满足Follow up)
+## 解题方案
 
-- 首先负数肯定不是palindrome
-- 其次如果一个数字是一个正数，并且能被我0整除那它肯定也不是palindrome
+> 思路1
 
-这样降低了复杂度
+* 排除小于0的数
+* 通过字符串进行反转，对比数字是否相等就行
 
 ```python
-class Solution(object):
+class Solution:
     def isPalindrome(self, x):
         """
         :type x: int
         :rtype: bool
         """
-        if x < 0 or (x != 0 and x % 10 == 0):
+        if x < 0:
             return False
-        rev, y = 0, x
-        while x > 0:
-            rev = rev * 10 + x % 10
-            x /= 10
-        return y == rev
-
+        elif x != int(str(x)[::-1]):
+            return False
+        else:
+            return True
+
+
+if __name__ == "__main__":
+    target = 12421
+    so = Solution()
+    status = so.isPalindrome(target)
+    print("结果: ", status)
 ```
 
+
 > 思路2
 
 * 排除小于0的数
-* 通过字符串进行反转，对比数字是否相等就行
+* 数字通过 % 10 求余数进行反转，进而得到最终反转的结果
+* 对比反转的结果是最初的结果是否相等
 
 ```python
-class Solution:
+class Solution(object):
     def isPalindrome(self, x):
         """
         :type x: int
         :rtype: bool
         """
         if x < 0:
             return False
-        elif x != int(str(x)[::-1]):
-            return False
-        else:
-            return True
+        rev, y = 0, x
+        while x > 0:
+            rev = rev * 10 + x % 10
+            print("余数: %s - 余数反转结果: %s" % (rev, x % 10))
+            x = int(x/10)
+        return y == rev
+
+
+if __name__ == "__main__":
+    target = 12321
+    so = Solution()
+    n = so.isPalindrome(target)
+    print("结果: ", n)
+"""
+余数: 1 - 余数反转结果: 1
+余数: 12 - 余数反转结果: 2
+余数: 123 - 余数反转结果: 3
+余数: 1232 - 余数反转结果: 2
+余数: 12321 - 余数反转结果: 1
+结果:  True
+"""
 ```