Description
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.)
解法一
思路：循环左移字符串k位与原来的字符串进行比较，（k位是由字符串长度len的因数决定）相等则为true。
如9因数有1，3，9，因为最大的substring为字符串长度的一半，所以
class Solution {
public:
string move(string s, int l) {
string res = s.substr(l);
res += s.substr(0,l);
return res;
}
bool repeatedSubstringPattern(string s) {
int len = s.size();
string str;
for (int i = 1; i <= len/2; i++) {
if (len % i == 0) {
str = move(s, i);
if(str == s) return true;
}
}
return false;
}
};Submission Details
107 / 107 test cases passed.
Status: Accepted
Runtime: 36 ms
解法二
用到KMP算法
1.Roughly speaking, dp[i+1] stores the maximum number of characters that the string is repeating itself up to position i.
2.Therefore, if a string repeats a length 5 substring 4 times, then the last entry would be of value 15.
3.To check if the string is repeating itself, we just need the last entry to be non-zero and str.size() to divide (str.size()-last entry).
bool repeatedSubstringPattern(string str) {
int i = 1, j = 0, n = str.size();
vector<int> dp(n+1,0);
while( i < str.size() ){
if( str[i] == str[j] ) dp[++i]=++j;
else if( j == 0 ) i++;
else j = dp[j];
}
return dp[n]&&dp[n]%(n-dp[n])==0;
}