刚刚又刷了一道看似简单，实则不然的题，整体感觉坑还行，下面来分享下做题的经验吧！
具体题目如下：
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".题意分析：
给定一个字符串s与一个非空字符串p，两字符串只包括小写字母，要求在s中找到所有包含p中所有字符且与p等大小的子串，并依次返回所有子串首元素的下标。
解答如下：
方法一（滑动窗口）
用p1去记录p中存放的元素，用s1去记录s中存放的前p个元素，通过不断滑动，再用等长度p1和s1去判断s1中所记录的元素是否包含p中元素。注：vector<int>p1(26,0); vector<int>s1(26,0); 这两句不可以用 p1[26] = {0}; s1[26] = {0}来替换。
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int> res;
vector<int>p1(26,0);
vector<int>s1(26,0);
if( s.size() < p.size() || p.size()==0) //为了避免s为空集，所以应该提前判断p.size()==0
return res;
for (int i = 0; i < p.size(); i++) //记录p中元素
p1[p[i] - 'a'] = p1[p[i] - 'a'] +1;
for (int i = 0; i < p.size(); i++) //记录s中前p个元素
s1[s[i] - 'a'] = s1[s[i] - 'a'] +1;
if (p1 == s1)
res.push_back(0);
for ( int i = 1; i <= s.size() - p.size(); i++) //形成一个个与p等大小的s子串，并作比较
{
s1[s[i-1]-'a']--;
s1[s[i-1+p.size()]-'a'] = s1[s[i-1+p.size()]-'a']+1;
if (p1 == s1)
res.push_back(i);
}
return res;
}
};
提交后的结果如下：
日积月累，与君共进，增增小结，未完待续。