tags: sliding window
Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
Example 1:
Input: s: "cbaebabacd" p: "abc"
Output: [0, 6]
Explanation: The substring with start index = 0 is "cba", which is an anagram of "abc". The substring with start index = 6 is "bac", which is an anagram of "abc". Example 2:
Input: s: "abab" p: "ab"
Output: [0, 1, 2]
Explanation: The substring with start index = 0 is "ab", which is an anagram of "ab". The substring with start index = 1 is "ba", which is an anagram of "ab". The substring with start index = 2 is "ab", which is an anagram of "ab".
// 32ms, 76%
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
int pbuf[26] = {0};
for (auto& c: p) pbuf[c-'a']++;
int sbuf[26] = {0};
int beg = 0;
vector<int> res;
for (int i = 0; i < s.size(); i++) {
int c = s[i] - 'a';
if (pbuf[c] < 1) {
while (beg < i) {
sbuf[s[beg]-'a']--;
beg++;
}
beg = i + 1;
} else {
sbuf[c]++;
if (i - beg + 1 == p.size()) {
if (memcmp(pbuf, sbuf, sizeof(sbuf)) == 0) {
res.push_back(beg);
}
sbuf[s[beg]-'a']--;
beg++;
}
}
}
return res;
}
};