459.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:
1
2
3
| Input: s = "abab"
Output: true
Explanation: It is the substring "ab" twice.
|
Example 2:
1
2
| Input: s = "aba"
Output: false
|
Example 3:
1
2
3
| Input: s = "abcabcabcabc"
Output: true
Explanation: It is the substring "abc" four times or the substring "abcabc" twice.
|
枚举法：
时间复杂度：O(n^2)
空间复杂度：O(1)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
| class Solution {
public:
bool repeatedSubstringPattern(string s) {
int n = s.size();
for (int i = 1; i * 2 <= n; ++i) {
if (n % i == 0) {
bool match = true;
for (int j = i; j < n; ++j) {
if (s[j] != s[j - i]) {
match = false;
break;
}
}
if (match) {
return true;
}
}
}
return false;
}
};
|
字符串匹配：
1
2
3
4
5
6
| class Solution {
public:
bool repeatedSubstringPattern(string s) {
return (s + s).find(s, 1) != s.size();
}
};
|
KMP算法：
时间复杂度：O(n)
空间复杂度：O(n)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
| class Solution {
public:
bool kmp(const string& query, const string& pattern) {
int n = query.size();
int m = pattern.size();
vector<int> fail(m, -1);
for (int i = 1; i < m; ++i) {
int j = fail[i - 1];
while (j != -1 && pattern[j + 1] != pattern[i]) {
j = fail[j];
}
if (pattern[j + 1] == pattern[i]) {
fail[i] = j + 1;
}
}
int match = -1;
for (int i = 1; i < n - 1; ++i) {
while (match != -1 && pattern[match + 1] != query[i]) {
match = fail[match];
}
if (pattern[match + 1] == query[i]) {
++match;
if (match == m - 1) {
return true;
}
}
}
return false;
}
bool repeatedSubstringPattern(string s) {
return kmp(s + s, s);
}
};
|
优化：
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
| class Solution {
public:
bool kmp(const string& pattern) {
int n = pattern.size();
vector<int> fail(n, -1);
for (int i = 1; i < n; ++i) {
int j = fail[i - 1];
while (j != -1 && pattern[j + 1] != pattern[i]) {
j = fail[j];
}
if (pattern[j + 1] == pattern[i]) {
fail[i] = j + 1;
}
}
return fail[n - 1] != -1 && n % (n - fail[n - 1] - 1) == 0;
}
bool repeatedSubstringPattern(string s) {
return kmp(s);
}
};
|