# 459. 重复的子字符串

2021/8/21 19:32:07 浏览：

## 代码

``````class Solution {
public boolean repeatedSubstringPattern(String s) {
String str = s + s;
//  return str.substring(1, str.length() - 1).contains(s);
return kmp(str.substring(1, str.length() - 1), s);
}

public boolean kmp(String haystack, String needle) {
int n = haystack.length(), m = needle.length();
if (m == 0) {
return false;
}
int[] pi = new int[m];
for (int i = 1, j = 0; i < m; i++) {
while (j > 0 && needle.charAt(i) != needle.charAt(j)) {
j=pi[j - 1];
}
if (needle.charAt(i) == needle.charAt(j)) {
j++;
}
pi[i]=j;
}
for (int i = 0, j = 0; i < n; i++) {
while (j > 0 && haystack.charAt(i) != needle.charAt(j)) {
j = pi[j - 1];
}
if (haystack.charAt(i) == needle.charAt(j)) {
j++;
}
if (j == m) {
return true;
}
}
return false;
}

}
``````

## 反思

1、以下是好的思路，调用库函数。

``````class Solution {
public boolean repeatedSubstringPattern(String s) {
String str = s + s;
return str.substring(1, str.length() - 1).contains(s);
//return kmp(str.substring(1, str.length() - 1), s);
}
}``````