leetcode30 串联所有单词的子串

思路：

滑动窗口+哈希表。

实现：

class Solution {
public:
    set<string>check(map<string,int>&cur,map<string,int>&mp){
        set<string>st;
        for(auto it:mp){
            string key=it.first;
            int val=it.second;
            if(!cur.count(key) or cur[key]!=val){
                st.insert(key);
            }
        }
        return st;
    }
    vector<int> findSubstring(string s, vector<string>& w) {
        int n=s.length(),m=w.size(),l=w[0].length();
        vector<int>res;
        map<string,int>mp;
        for(auto it:w){
            mp[it]++;
        }
        for(int i=0;i<l;i++){
            map<string,int>cur;
            int cnt=0;
            int j=i;
            for(;j+l-1<n and cnt<m;j+=l){
                string tmp=s.substr(j,l);
                cur[tmp]++;
                cnt++;
            }
            if(cnt<m){
                break;
            }
            set<string>diff=check(cur,mp);
            if(diff.empty()){
               res.push_back(j-l*m);
            }
            for(;j+l-1<n;j+=l){
                string tmp=s.substr(j,l);
                string last=s.substr(j-m*l,l);
                cur[tmp]++;
                cur[last]--;
                if(cur[last]==0){
                    cur.erase(last);
                }
                if(!mp.count(tmp) or cur[tmp]!=mp[tmp]){
                    diff.insert(tmp);
                }
                else{
                    diff.erase(tmp);
                }
                if(cur.count(last)){

                    if(!mp.count(last) or cur[last]!=mp[last]){
                        diff.insert(last);
                    }
                    else{
                        diff.erase(last);
                    }
                }
                else{
                    if(mp.count(last)){
                        diff.insert(last);
                    }
                    else{
                        diff.erase(last);
                    }
                }
                if(diff.empty()){
                    res.push_back(j-l*(m-1));
                }
            }
        }
        return res; 
    }
};