459. Repeated Substring Pattern - Easy

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.)

 

重复子串的长度一定是字符串长度的约数，从len/2开始查找，如果找到了能被整除的，len/i = m，把字符串从0~i这一“重复” pattern append m次，看是否与原字符串相同。如果遍历完所有约数没有结果，返回false

时间：O(N^2)，空间O(N)

class Solution {
    public boolean repeatedSubstringPattern(String s) {
        for(int i = 1; i < s.length(); i++) {
            if(s.length() % i == 0) {
                int m = s.length() / i;
                StringBuilder sb = new StringBuilder();
                for(int j = 0; j < m; j++) {
                    sb.append(s.substring(0, i));
                }
                if(sb.toString().equals(s))
                    return true;
            }
        }
        return false;
    }
}

 

改进：

时间：worst case O(n^2), average O(n) ~ O(n^2)，空间：O(1)

class Solution {
    public boolean repeatedSubstringPattern(String s) {
        for(int i = s.length() / 2; i >= 1 ; i--) {
            if(s.length() % i == 0) {
                int m = s.length() / i;
                String sb = s.substring(0, i);
                int j;
                for(j = 1; j < m; j++) {
                    if(!sb.equals(s.substring(j*i,i+j*i)))
                        break;
                }
                if(j == m)
                    return true;
            }
        }
        return false;
    }
}