For a string sequence, a string word is k-repeating if word concatenated k times is a substring of sequence. The word's maximum k-repeating value is the highest value k where word is k-repeating in sequence. If word is not a substring of sequence, word's maximum k-repeating value is 0.
Given strings sequence and word, return the maximum k-repeating value of word in sequence.
Example 1
Input: sequence = "ababc", word = "ab"
Output: 2
Explanation: "abab" is a substring in "ababc".
Example 2
Input: sequence = "ababc", word = "ba"
Output: 1
Explanation: "ba" is a substring in "ababc". "baba" is not a substring in "ababc".
Example 3
Input: sequence = "ababc", word = "ac"
Output: 0
Explanation: "ac" is not a substring in "ababc".
Constraints
- 1 <= sequence.length <= 100
- 1 <= word.length <= 100
- sequence and word contains only lowercase English letters.
Solution
默写 KMP
class Solution {
public:
void buildMatch(vector<int> &match, string pattern) {
match[0] = -1;
for (int i = 1; i < match.size(); i++) {
int j = match[i - 1];
while (j >= 0 && pattern[i] != pattern[j + 1])
j = match[j];
if (pattern[i] == pattern[j + 1]) match[i] = j + 1;
else match[i] = -1;
}
}
int KMP(string sstring, string pattern) {
int n = sstring.length();
int m = pattern.length();
if (n < m) return -1;
vector<int> match(m);
buildMatch(match, pattern);
int s, p;
s = p = 0;
while (s < n && p < m) {
if (sstring[s] == pattern[p]) {
s++; p++;
} else if (p > 0)
p = match[p - 1] + 1;
else
s++;
}
return (p == m) ? s - m : -1;
}
int maxRepeating(string sequence, string word) {
int ans = 0;
string pattern = word;
while (KMP(sequence, pattern) != -1) {
ans++;
pattern += word;
}
return ans;
}
};
