From b8bb7786a41a78bca103f686398fd2801e43bdbb Mon Sep 17 00:00:00 2001
From: hk27xing <244798299@qq.com>
Date: Wed, 19 May 2021 16:23:04 +0800
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---
 problems/0459.重复的子字符串.md | 31 +++++++++++++++++++++++++-
 1 file changed, 30 insertions(+), 1 deletion(-)

diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md
index 5abc999a..3a5f3b17 100644
--- a/problems/0459.重复的子字符串.md
+++ b/problems/0459.重复的子字符串.md
@@ -149,9 +149,38 @@ public:
 
 ## 其他语言版本
 
-
 Java：
 
+```java
+class Solution {
+    public boolean repeatedSubstringPattern(String s) {
+        if (s.equals("")) return false;
+
+        int len = s.length();
+        // 原串加个空格(哨兵)，使下标从1开始，这样j从0开始，也不用初始化了
+        s = " " + s;
+        char[] chars = s.toCharArray();
+        int[] next = new int[len + 1];
+
+        // 构造 next 数组过程，j从0开始(空格)，i从2开始
+        for (int i = 2, j = 0; i <= len; i++) {
+            // 匹配不成功，j回到前一位置 next 数组所对应的值
+            while (j > 0 && chars[i] != chars[j + 1]) j = next[j];
+            // 匹配成功，j往后移
+            if (chars[i] == chars[j + 1]) j++;
+            // 更新 next 数组的值
+            next[i] = j;
+        }
+
+        // 最后判断是否是重复的子字符串，这里 next[len] 即代表next数组末尾的值
+        if (next[len] > 0 && len % (len - next[len]) == 0) {
+            return true;
+        }
+        return false;
+    }
+}
+```
+
 
 Python：