30.查找所有可能的字符串组合

Substring with Concatenation of All Words

问题描述：

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).

知识补充：

unordered_map

unordered_map<int, int> c;//<type,type>分别对应键的类型和键值的类型c[0] = 1;//简单的插入数据形式c.find(key)==c.end();//判断无序图内是否有这个键c.count(key);//访问键值

测试代码：

    vector<int> findSubstring(string s, vector<string>& words) {        vector<int> result;        unordered_map<string, int> c;        unordered_map<string, int> n;        int j,i,k;        string ss;        int l = words[0].length();        for (string word : words)            c[word]++;        for(i=0;i<s.length()-l*words.size()+1;i++)        {            j=i;            k=words.size();                        n = c;            while(k)            {                ss = s.substr(j,l);                if(n[ss]<1)                {                    break;                }                if(n.find(ss)!=n.end())                {                    --n[ss];                    j = j+l;                    k--;                }else{                    break;                }            }            if(k==0)            {                result.push_back(i);            }        }        return result;    }

性能：

参考答案：

class Solution {public:    vector<int> findSubstring(string s, vector<string>& words) {        vector<int> ret;        if(words.empty() || words[0].length()==0) return ret;        int wl=words[0].length();        int len=wl*words.size();        int end=s.length()-len;        if(end<0) return ret;        unordered_map<string, int> dict;        for(string word : words) dict[word]++;        for(int st=0; st<wl; st++) {            helper(ret, dict, s, words, st, end, wl, len);        }        return ret;    }private:    void helper(vector<int>& ret, unordered_map<string, int>& dict, string& s, vector<string>& words, int st, int end, int wl, int len) {        unordered_map<string, int> hash;        int left=st, right=st;        while(left<=end) {            string nxt=s.substr(right, wl);            if(dict.count(nxt)==0) {                left=right+wl;                hash.clear();            }            else {                if(hash.count(nxt)==0) hash[nxt]=1;                else if(hash[nxt]<dict[nxt]) hash[nxt]++;                else {                    string rem=s.substr(left, wl);                    while(rem!=nxt) {                        hash[rem]--;                        left+=wl;                        rem=s.substr(left, wl);                    }                    left+=wl;                }            }            right+=wl;            if(right-left==len) ret.push_back(left);        }    }};

性能：