solve problem Reverse Integer

Author: P-ppc

README.md

@@ -184,6 +184,7 @@ All solutions will be accepted!
 |581|[Shortest Unsorted Continuous Subarray](https://leetcode-cn.com/problems/shortest-unsorted-continuous-subarray/description/)|[java/py/js](./algorithms/ShortestUnsortedContinuousSubarray)|Easy|
 |443|[String Compression](https://leetcode-cn.com/problems/string-compression/description/)|[java/py/js](./algorithms/StringCompression)|Easy|
 |438|[Find All Anagrams In A String](https://leetcode-cn.com/problems/find-all-anagrams-in-a-string/description/)|[java/py/js](./algorithms/FindAllAnagramsInAString)|Easy|
+|7|[Reverse Integer](https://leetcode-cn.com/problems/reverse-integer/description/)|[java/py/js](./algorithms/ReverseInteger)|Easy|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/ReverseInteger/README.md

@@ -0,0 +1,2 @@
+# Reverse Integer
+This problem is easy to solve

algorithms/ReverseInteger/Solution.java

@@ -0,0 +1,19 @@
+class Solution {
+    public int reverse(int x) {
+        int res = 0,
+            MAX_INTEGER_BOUNDARY = Integer.MAX_VALUE / 10,
+            MIN_INTEGER_BOUNDARY = Integer.MIN_VALUE / 10;
+        
+        while (x != 0) {
+            int pop = x % 10;
+            x /= 10;
+            
+            if (res > MAX_INTEGER_BOUNDARY || (res == MAX_INTEGER_BOUNDARY && pop > 7)) return 0;
+            if (res < MIN_INTEGER_BOUNDARY || (res == MIN_INTEGER_BOUNDARY && pop < -8)) return 0;
+            
+            res = res * 10 + pop;
+        }
+        
+        return res;
+    }
+}

algorithms/ReverseInteger/solution.js

@@ -0,0 +1,20 @@
+/**
+ * @param {number} x
+ * @return {number}
+ */
+var reverse = function(x) {
+    let res = 0,
+        MAX_INTEGER_BOUNDARY = parseInt((Math.pow(2, 31) - 1) / 10),
+        MIN_INTEGER_BOUNDARY = parseInt(Math.pow(-2, 31) / 10)
+    
+    while (x != 0) {
+        let pop = x % 10
+        x = parseInt(x / 10)
+        
+        if (res > MAX_INTEGER_BOUNDARY || (res === MAX_INTEGER_BOUNDARY && pop > 7)) return 0
+        if (res < MIN_INTEGER_BOUNDARY || (res === MIN_INTEGER_BOUNDARY && pop < -8)) return 0
+        res = res * 10 + pop
+    }
+    
+    return res
+};

algorithms/ReverseInteger/solution.py

@@ -0,0 +1,16 @@
+class Solution(object):
+    def reverse(self, x):
+        """
+        :type x: int
+        :rtype: int
+        """
+        res = 0
+        is_negative = x < 0
+        x = abs(x)
+        
+        while x > 0:
+            res = res * 10 + x % 10
+            x /= 10
+        
+        res = res if not is_negative else - res
+        return res if pow(-2, 31) <= res <= pow(2, 31) - 1 else 0