题目描述：
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".
要完成的函数：
vector<int> findAnagrams(string s, string p)
说明：
1、给定一个字符串s和非空字符串p，将p中元素不断交换形成一个新的字符串，如果这个新的字符串在s中能找到匹配的，那么就输出匹配到的开始的位置，直到处理到字符串s结束。
2、这道题目难道要记住p经过交换可能形成的所有字符串吗，难道再类似于滑动窗口一般不断在s中比较？
其实不用记住所有字符串，记住p经过交换可能形成的所有字符串其实等价于记住p中所有字母出现的次数。
所以代码如下：
vector<int> findAnagrams(string s, string p)
{
vector<int>res;
vector<int>p1(26,0);
vector<int>s1(26,0);
for(int i=0;i<p.size();i++)
{
p1[p[i]-'a']++;
}
for(int i=0;i<p.size();i++)
{
s1[s[i]-'a']++;
}
if(s.size()<p.size())
return res;
if(p1==s1)
res.push_back(0);
for(int i=1;i<=s.size()-p.size();i++)
{
s1[s[i-1]-'a']--;//当窗口滑动时，s1中要减去窗口前一位的字母1次
s1[s[i-1+p.size()]-'a']++;//s1要加上窗口最后新的一维的字母1次
if(s1==p1)
res.push_back(i);
}
return res;
}
上述代码实测35ms，beats 90.84% of cpp submissions。