add leetcode9 js solution

Author: swiftwind0405

README.md

@@ -433,7 +433,7 @@
 |12|[Integer to Roman](https://leetcode.com/problems/integer-to-roman/)| |Medium|
 |11|[Container With Most Water](https://leetcode.com/problems/container-with-most-water/)| [js](./algorithms/containerWithMostWater/containerWithMostWater.js) |Medium|
 |10|[Regular Expression Matching](https://leetcode.com/problems/regular-expression-matching/)| |Hard|
-|9|[Palindrome Number](https://leetcode.com/problems/palindrome-number/)| |Easy|
+|9|[Palindrome Number](https://leetcode.com/problems/palindrome-number/)| [js](./algorithms/palindromeNumber/Solution.js) |Easy|
 |8|[String to Integer (atoi)](https://leetcode.com/problems/string-to-integer-atoi/)| |Easy|
 |7|[Reverse Integer](https://leetcode.com/problems/reverse-integer/)| [js](./algorithms/reverseInteger/reverseInteger.js),[java](./algorithms/reverseInteger/Solution.java)  |Easy|
 |6|[ZigZag Conversion](https://leetcode.com/problems/zigzag-conversion/)| |Easy|

algorithms/palindromeNumber/Solution.js

@@ -0,0 +1,45 @@
+// 普通解法：数字转为字符串，然后将字符串分割为数组，只需要循环数组的一半长度进行判断对应元素是否相等即可。
+ var isPalindrome1 = function(x) {
+    // 特殊情况：
+    // 如上所述，当 x < 0 时，x 不是回文数。
+    // 同样地，如果数字的最后一位是 0，为了使该数字为回文，
+    // 则其第一位数字也应该是 0
+    // 只有 0 满足这一属性
+    if (x < 0 || (x % 10 === 0 && x !== 0)) {
+        return false;
+    }
+
+    const xStr = x.toString();
+    const length = xStr.length;
+    const middle = length / 2;
+    for (let x = 0, y = length - 1; x <= middle; x++ , y--) {
+        if (xStr[x] !== xStr[y]) {
+            return false;
+        }
+    }
+    return true;
+};
+
+// 数学解法：通过取整和取余操作获取整数中对应的数字进行比较。
+var isPalindrome2 = function(x) {
+    if (x < 0) {
+        return false;
+    }
+
+    let div = 1;
+    while (x / div >= 10) {
+        div *= 10
+    }
+    console.log('div', div)
+    while (x > 0) {
+        let left = parseInt(x / div),
+            right = x % 10;
+        if (left !== right) {
+            return false
+        }
+        // 再往中间进一位继续判断
+        x = parseInt((x % div) / 10);
+        div /= 100
+    }
+    return true;
+};