leetcode-java/src/RepeatedSubstringPattern459.java at master · fluency03/leetcode-java · GitHub

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/**
 * Given a non-empty string check if it can be constructed by taking a
 * substring of it and appending multiple copies of the substring together.
 * You may assume the given string consists of lowercase English letters only
 * and its length will not exceed 10000.
 *
 * Example 1:
 * Input: "abab"
 * Output: True
 * Explanation: It's the substring "ab" twice.
 *
 * Example 2:
 * Input: "aba"
 * Output: False
 *
 * Example 3:
 * Input: "abcabcabcabc"
 * Output: True
 * Explanation: It's the substring "abc" four times. (And the substring "abcabc" twice.)
 */

publicclassRepeatedSubstringPattern459 {

/**
 * https://leetcode.com/problems/repeated-substring-pattern/discuss/94352/Java-Simple-Solution-with-Explanation
 */
publicbooleanrepeatedSubstringPattern(Strings) {
if(s==null || s.length()<=1) returnfalse;

for(inti=1;i<=s.length()/2;i++){
if(s.length()%i!=0) continue;
Stringsub = s.substring(0,i);
if(dfs(s,sub,i)) returntrue;
 }
returnfalse;
 }

privatebooleandfs(Strings,Stringsub,inti){
if(i==s.length()) returntrue;
if(!s.startsWith(sub,i)) returnfalse;
returndfs(s,sub,i+sub.length());
 }


/**
 * https://leetcode.com/problems/repeated-substring-pattern/discuss/94344/Simple-Java-solution-2-lines
 */
publicbooleanrepeatedSubstringPattern2(Stringstr) {
Strings = str + str;
returns.substring(1, s.length() - 1).contains(str);
 }


/**
 * https://leetcode.com/problems/repeated-substring-pattern/discuss/94340/Java-and-O(n)
 */
publicbooleanrepeatedSubstringPattern3(Stringstr) {
//This is the kmp issue
int[] prefix = kmp(str);
intlen = prefix[str.length()-1];
intn = str.length();
return (len > 0 && n%(n-len) == 0);
 }

privateint[] kmp(Strings){
intlen = s.length();
int[] res = newint[len];
char[] ch = s.toCharArray();
inti = 0, j = 1;
res[0] = 0;
while(i < ch.length && j < ch.length){
if(ch[j] == ch[i]){
res[j] = i+1;
i++;
j++;
 }else{
if(i == 0){
res[j] = 0;
j++;
 }else{
i = res[i-1];
 }
 }
 }
returnres;
 }

}