Ideas :
(1) Sliding window
The core idea of the sliding window here is to first intercept the first substring whose length is a short string in the long string, then move the right pointer to the right ++, and the left pointer to the right to –, which respectively represent joining A new number and reduce an old number, so that the substring can be found.
class Solution {
public:
bool checkInclusion(string s1, string s2) {
int n = s1.length(), m = s2.length();
if (n > m) {
return false;
}
vector<int> cnt1(26), cnt2(26);
for (int i = 0; i < n; ++i) {
++cnt1[s1[i] - 'a'];
++cnt2[s2[i] - 'a'];
}
if (cnt1 == cnt2) {
return true;
}
for (int i = n; i < m; ++i) {
++cnt2[s2[i] - 'a'];
--cnt2[s2[i - n] - 'a'];
if (cnt1 == cnt2) {
return true;
}
}
return false;
}
};
Idea:
It is similar to the above idea. When you find a match, don’t return directly and use a vector to save it.
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
int n = p.length(), m = s.length();
vector<int>res;
if (n > m) {
return res;
}
vector<int> cnt1(26), cnt2(26);
for (int i = 0; i < n; ++i) {
++cnt1[p[i] - 'a'];
++cnt2[s[i] - 'a'];
}
if (cnt1 == cnt2) {
res.push_back(0);
}
for (int i = n; i < m; ++i) {
++cnt2[s[i] - 'a'];
--cnt2[s[i - n] - 'a'];
if (cnt1 == cnt2) {
res.push_back(i-n+1);
}
}
return res;
}
};