力扣链接：459. 重复的子字符串，难度等级：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 <= 10000sconsists of lowercase English letters.
复杂度
时间复杂度
O(N * N)
空间复杂度
O(N)
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