feat: daily 2019-07-25 9. Palindrome Number (azl397985856#66)

* feat: longest harmonious subsequence

* daily: Longest Harmonious Subsequence, option 1

* pref: make code universal.

* pref: get better answer.

* alternative solutions.

* Update 2019-06-20.md

* Update 2019-06-20.md

* feat: date 2019.06.17

* fix: add tag.

* Update 2019-06-17.md

* Update 2019-06-17.md

* feat: 9.palindrome-number(2019-07-25)

Author: EdgeAsh

daily/2019-07-25.md

@@ -0,0 +1,92 @@
+# 毎日一题 - 9. Palindrome number
+
+## 信息卡片
+
+- 时间：2019-07-25
+- 题目链接：https://leetcode.com/problems/palindrome-number/submissions/
+- tag：`Math`
+
+## 题目描述
+
+```
+Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
+
+Example 1:
+
+  Input: 121
+  Output: true
+
+Example 2:
+
+  Input: -121
+  Output: false
+  Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
+
+Example 3:
+
+  Input: 10
+  Output: false
+  Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
+
+Follow up:
+
+  Coud you solve it without converting the integer to a string?
+```
+
+## 参考答案
+
+转成字符串方式
+1. 负数都是非回文数，10的整数倍不是回文。
+2. 将数字转为字符串，再逆序排列字符串。两者比较，相等就是回文数。
+
+直接操作整数方式
+1. 复制x到temp;
+2. 取temp末尾数字，方式为temp与10的求余；组成新数reverse;
+3. 每取完一位,temp缩小10倍并且去掉小数。
+4. reverse要`先扩大十倍`再加上取下来的数
+5. 当temp === 0时，表示已经取完；reverse与x比较
+
+
+参考JavaScript代码：
+
+```js
+/**
+ * @param {number} x
+ * @return {boolean}
+ * 转成字符串
+ */
+var isPalindrome = function(x) {
+    if(x < 0 ) return false;
+    if(x === 0) return true;
+    if(x % 10 === 0) return false;
+    let reverse = '';
+    let str = String(x);
+    for(let i = str.length - 1; i >= 0; i--) {
+      reverse += str[i]
+    }
+    return str === reverse
+};
+
+/**
+ * @param {number} x
+ * @return {boolean}
+ * 不转成字符串
+ */
+var isPalindrome = function(x) {
+    if(x < 0 ) return false;
+    if(x === 0) return true;
+    if(x % 10 === 0) return false;
+    let temp = x;
+    let reverse = 0;
+    while(temp > 0) {
+      let num = temp % 10;
+      temp = (temp - num)/10; // 或 temp = (temp / 10) >> 0,去除小数位
+      reverse = reverse * 10 + num
+    }
+    return x === reverse
+};
+```
+
+## 优秀解答
+
+> 暂缺

daily/README.md

@@ -155,3 +155,9 @@ tag：`Linked List`
 tag：`发散思维`
 
 时间: 2019-07-24
+
+### [9.palindrome-number](./2019-07-25.md)
+
+tag：`Math`
+
+时间: 2019-07-25