solve problem Repeated Substring Pattern

Author: P-ppc

README.md

@@ -179,6 +179,7 @@ All solutions will be accepted!
 |189|[Rotate Array](https://leetcode-cn.com/problems/rotate-array/description/)|[java/py/js](./algorithms/RotateArray)|Easy|
 |687|[Longest Univalue Path](https://leetcode-cn.com/problems/longest-univalue-path/description/)|[java/py/js](./algorithms/LongestUnivaluePath)|Easy|
 |28|[Implement Strstr](https://leetcode-cn.com/problems/implement-strstr/description/)|[java/py/js](./algorithms/ImplementStrstr)|Easy|
+|459|[Repeated Substring Pattern](https://leetcode-cn.com/problems/repeated-substring-pattern/description/)|[java/py/js](./algorithms/RepeatedSubstringPattern)|Easy|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/RepeatedSubstringPattern/README.md

@@ -0,0 +1,2 @@
+# Repeated Substring Pattern
+We can use KMP to solve this problem

algorithms/RepeatedSubstringPattern/Solution.java

@@ -0,0 +1,21 @@
+class Solution {
+    public boolean repeatedSubstringPattern(String s) {
+        int length = s.length(),
+            j = -1;
+        int[] next = new int[length];
+        next[0] = -1;
+        
+        for (int i = 1; i < length; i++) {
+            while (j >= 0 && s.charAt(i) != s.charAt(j + 1)) {
+                j = next[j];
+            }
+            if (s.charAt(i) == s.charAt(j + 1)) {
+                j++;
+            }
+            next[i] = j;
+        }
+        
+        int subLength = length - 1 - next[length - 1];
+        return subLength != length && length % subLength == 0;
+    }
+}

algorithms/RepeatedSubstringPattern/solution.js

@@ -0,0 +1,24 @@
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var repeatedSubstringPattern = function(s) {
+    let i = j = 0,
+        next = []
+    
+    while (i < s.length) {
+        if (i === 0 || (s[i] !== s[j] && j === 0)) {
+            next.push(0)
+        } else if (s[i] === s[j]) {
+            next.push(j + 1)
+            j++
+        } else if (s[i] !== s[j]) {
+            j = next[j - 1]
+            continue
+        }
+        i++
+    }
+    
+    let subLength = s.length - next[s.length - 1]
+    return subLength !== s.length && s.length % subLength === 0
+};

algorithms/RepeatedSubstringPattern/solution.py

@@ -0,0 +1,25 @@
+class Solution(object):
+    def repeatedSubstringPattern(self, s):
+        """
+        :type s: str
+        :rtype: bool
+        """
+        # method of exhaustion
+        for length in range(1, len(s) / 2 + 1):
+            if len(s) % length != 0:
+                continue
+                
+            start = 0
+            temp = None
+            is_equal = True
+            while start < len(s):
+                p = s[start:start + length]
+                if temp == None:
+                    temp = p
+                elif temp != p:
+                    is_equal = False
+                    break
+                start += length
+            if is_equal:
+                return True
+        return False