substring-with-concatenation-of-all-words

题目:

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.
For example, given:
S:”barfoothefoobarman”
L:[“foo”, “bar”]
You should return the indices:[0,9].
(order does not matter).

程序：

class Solution {public:    vector<int> findSubstring(string S, vector<string> &L) {        map<string, int> dict;        vector<int> res;        if (!S.size() || !L.size())            return res;        for (int i = 0; i < L.size(); i++)            dict[L[i]] ++;        int m = L.size();        int n = L[0].size();        int size = m*n;        for (int i = 0; i < S.size() - size + 1; i++)        {            for (int a = 0; a < m; a++)//字典归0            {                dict[L[a]] = 0;            }            for (int a = 0; a < m; a++)//与上步一起初识化字典            {                dict[L[a]] ++;            }            string s = S.substr(i, size);            int j = 0;            while (j < m)            {                string ss = s.substr(j*n, n);                if (dict.find(ss) != dict.end())                    dict[ss]--;                else                    break;                j++;            }            if (j != m)                continue;            int k = 0;            for (; k < m; k++)            {                if (dict[L[k]])                    break;            }            if (k == m)                res.push_back(i);        }        return res;    }};

点评：

通过哈希表减少时间复杂度