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;//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;i1;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; streturn 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]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：