提供JavaScript版本的《重复的子字符串》

Author: kok-s0s

problems/0459.重复的子字符串.md

@@ -289,8 +289,79 @@ func repeatedSubstringPattern(s string) bool {
 }
 ```
 
+JavaScript版本
 
+> 前缀表统一减一
 
+```javascript
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var repeatedSubstringPattern = function (s) {
+    if (s.length === 0)
+        return false;
+
+    const getNext = (s) => {
+        let next = [];
+        let j = -1;
+
+        next.push(j);
+
+        for (let i = 1; i < s.length; ++i) {
+            while (j >= 0 && s[i] !== s[j + 1])
+                j = next[j];
+            if (s[i] === s[j + 1])
+                j++;
+            next.push(j);
+        }
+
+        return next;
+    }
+
+    let next = getNext(s);
+
+    if (next[next.length - 1] !== -1 && s.length % (s.length - (next[next.length - 1] + 1)) === 0)
+        return true;
+    return false;
+};
+```
+
+> 前缀表统一不减一
+
+```javascript
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var repeatedSubstringPattern = function (s) {
+    if (s.length === 0)
+        return false;
+
+    const getNext = (s) => {
+        let next = [];
+        let j = 0;
+
+        next.push(j);
+
+        for (let i = 1; i < s.length; ++i) {
+            while (j > 0 && s[i] !== s[j])
+                j = next[j - 1];
+            if (s[i] === s[j])
+                j++;
+            next.push(j);
+        }
+
+        return next;
+    }
+
+    let next = getNext(s);
+
+    if (next[next.length - 1] !== 0 && s.length % (s.length - next[next.length - 1]) === 0)
+        return true;
+    return false;
+};
+```