leetcode-459-easy

Repeated Substring Pattern

Given a string s, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.

Example 1:

Input: s = "abab"
Output: true
Explanation: It is the substring "ab" twice.
Example 2:

Input: s = "aba"
Output: false
Example 3:

Input: s = "abcabcabcabc"
Output: true
Explanation: It is the substring "abc" four times or the substring "abcabc" twice.
Constraints:

1 <= s.length <= 104
s consists of lowercase English letters.

思路一：长度为 n 的字符串，获取字符串的子字符串，长度从[1, n/2]，然后从这个子字符串构造完整的字符串，最后判断两者是否匹配。

    public boolean repeatedSubstringPattern(String s) {
        for (int i = 1; i <= s.length() / 2; i++) {
            String subStr = s.substring(0, i);
            if (repeatedStringMatch(subStr, s)) {
                return true;
            }
        }

        return false;
    }

    public static boolean repeatedStringMatch(String subStr, String s) {
        int subLength = subStr.length();
        int strLength = s.length();

        if (strLength % subLength != 0) {
            return false;
        }

        StringBuilder sb = new StringBuilder();
        for (int i = 1; i <= strLength / subLength; i++) {
            sb.append(subStr);
        }

        return sb.toString().equals(s);
    }

思路二：看了官方题解，发现有一种更巧妙的解题思路

假设字符串s是由s1+s2组成的，s+s后，str就变成了s1+s2+s1+s2，去掉首尾，破环了首尾的s1和s2，变成了s3+s2+s1+s4,此时str中间就是s2+s1,如果s是循环字串，也就是s1=s2,所以str中间的s2+s1就和原字符串相等。如果s不是循环字串，s1!=s2，那么s1+s2是不等于s2+s1的，也就是str中间不包含s

        String str = s + s;
        return str.substring(1, str.length() - 1).contains(s);