1 map 固定窗口实现 不够速度 词频统计
class Solution { public:vector<int> findAnagrams(string s, string p) {map<char,int> p_map;map<char,int> s_map;vector<int> result_;for(int i=0;i<p.size();i++){if(p_map.contains(p[i])){p_map[p[i]]++;}else{p_map[p[i]]=1;}}// p 6 s 10 i 4(0) i=5 L=i-0+1=6 for(int i=0;i<s.size();i++){if(s_map.contains(s[i])){s_map[s[i]]++;}else{s_map[s[i]]=1;}int left=i-p.size()+1;if(left<0){continue;}if(s_map==p_map){result_.push_back(left);}s_map[s[left]]--;if( s_map[s[left]]==0){s_map.erase(s[left]);}}return result_;}};