【30】Substring with Concatenation of All Words

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

用一个map记录words中每个词出现的次数，然后遍历s，判断从s的每一个位置能否构成words的一个组合。判断的时候用另一个map记录s中每个词出现的次数，如果某个词出现的次数大于在words中出现的次数或某个词不在words中，则此位置不能构成words的一个组合，结束本次循环并进入下一个位置的判断

vector<int> findSubstring(string s, vector<string>& words) {    map< string , int > allwords;    map< string , int > wcount;    vector<int> res;    int len=s.length();    int n=words.size();    if(n==0)return res;    int wl=words[0].length();    for(int i=0;i<n;i++){        if(allwords.find(words[i])==allwords.end()){            allwords.insert(pair<string,int>(words[i],1));        }        else allwords[words[i]]=allwords[words[i]]+1;    }    for(int i=0;i<=len-wl*n;i++){        int wnum=0;        wcount.clear();        for(int j=0;j<n;j++){            string cs=s.substr(i+j*wl,wl);            if(allwords.find(cs)!=allwords.end()){                if(wcount.find(cs)!=wcount.end()){                    wcount[cs]=wcount[cs]+1;                    if(wcount[cs]>allwords[cs])break;                    wnum++;                }                else{                    wcount.insert(pair<string,int>(cs,1));                    wnum++;                }                if(wnum==n)res.push_back(i);            }            else break;        }    }    return res;}