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".滑动窗口,每次只有首尾字母变化，统计变化即可
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int> res,pCount(26,0),sCount(26,0);
int m=s.size();
int n=p.size();
if(n>m)
return res;
if(n==0)
{
for(int i=0;i<m;i++)
res.push_back(i);
}
for(int i=0;i<n;i++)
{
pCount[p[i]-'a']++;
sCount[s[i]-'a']++;
}
if(pCount==sCount)
res.push_back(0);
for(int j=n;j<m;j++)
{
sCount[s[j]-'a']++;//滑动窗口每次只有一个字母变化，对窗口首尾的变化统计即可
sCount[s[j-n]-'a']--;
if(pCount==sCount)
res.push_back(j-n+1);
}
return res;
}
};