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    力扣链接：459. 重复的子字符串，难度：简单。

    给定一个非空的字符串 s ，检查是否可以通过由它的一个子串重复多次构成。

    示例 1:

    输入: s = "abcabcabcabc"

    输出: true

    解释: 可由子串 "abc" 重复四次构成。 (或子串 "abcabc" 重复两次构成。)

    示例 2:

    输入: s = "aba"

    输出: false

    约束:

    • 1 <= s.length <= 10000
    • s consists of lowercase English letters.

    思路
    上一题下一题

    解决本问题的关键是要看清楚一点：通过子串的重复能得到s，那么子串的起始字母一定是s[0]。
    想明白了这一点，子串的排查范围就大大缩小了。

    复杂度

    时间复杂度

    O(N * N)

    空间复杂度

    O(N)

    题解语言： Python JavaScript Go C++ Java C# Ruby

    Python # 

    class Solution:
    def repeatedSubstringPattern(self, s: str) -> bool:
        for i in range(1, int(len(s) / 2) + 1):
            if len(s) % i == 0 and s[:i] * int(len(s) / i) == s:
                return True

        return False

    JavaScript # 

    var repeatedSubstringPattern = function (s) {
  for (let i = 1; i <= s.length / 2; i++) {
    if (s.length % i != 0) {
      continue
    }

    if (s.slice(0, i).repeat(s.length / i) == s) {
      return true
    }
  }

  return false
};

    Go # 

    func repeatedSubstringPattern(s string) bool {
    n := len(s)

    for i := 1; i <= n/2; i++ {
        if n%i == 0 {
            substring := s[:i]
            repeated := strings.Repeat(substring, n/i)

            if repeated == s {
                return true
            }
        }
    }

    return false
}

    C++ # 

    class Solution {
public:
    bool repeatedSubstringPattern(string s) {
        int n = s.length();

        for (int i = 1; i <= n / 2; i++) {
            if (n % i != 0) {
                continue;
            }

            string pattern = s.substr(0, i);
            string repeated = "";

            for (int j = 0; j < n / i; j++) {
                repeated += pattern;
            }

            if (repeated == s) {
                return true;
            }
        }

        return false;
    }
};

    Java # 

    class Solution {
    public boolean repeatedSubstringPattern(String s) {
        int n = s.length();

        for (var i = 1; i <= n / 2; i++) {
            if (n % i != 0) {
                continue;
            }

            var pattern = s.substring(0, i);
            var repeated = new StringBuilder();

            // Simply concatenate the pattern multiple times
            for (var j = 0; j < n / i; j++) {
                repeated.append(pattern);
            }

            // Compare the constructed string with the original string
            if (repeated.toString().equals(s)) {
                return true;
            }
        }

        return false;
    }
}

    C# # 

    public class Solution
{
    public bool RepeatedSubstringPattern(string s)
    {
        int n = s.Length;

        for (int i = 1; i <= n / 2; i++)
        {
            if (n % i != 0)
            {
                continue;
            }

            // Get the potential substring pattern
            string pattern = s.Substring(0, i);
            StringBuilder repeated = new StringBuilder();

            // Simply concatenate the pattern multiple times
            for (int j = 0; j < n / i; j++)
            {
                repeated.Append(pattern);
            }

            // Compare the constructed string with the original string
            if (repeated.ToString() == s)
            {
                return true;
            }
        }

        return false;
    }
}

    Ruby # 

    # @param {String} s
# @return {Boolean}
def repeated_substring_pattern(s)
  n = s.length

  (1..n / 2).each do |i|
    next unless n % i == 0

    pattern = s[0...i]
    repeated = ""

    # Simply concatenate the pattern multiple times
    (0...(n / i)).each do
      repeated += pattern
    end

    # Compare the constructed string with the original string
    return true if repeated == s
  end

  false
end
    题解语言： Python JavaScript Go C++ Java C# Ruby

    其它语言

    欢迎贡献代码到我们的 GitHub 仓库，非常感谢！本题解位置在 459. 重复的子字符串。
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