LeetCode30. Substring with Concatenation of All Words

题意：

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.

s: “barfoothefoobarman”
words: [“foo”, “bar”]
You should return the indices: [0,9]. (order does not matter).

words里面是可能重复的。

初始做法，使用多重映射multimap，很朴素的做法，超时：

class Solution {public:    vector<int> findSubstring(string s, vector<string>& words) {        vector<int> vec;        if(words.empty()) return vec;        int num = words.size();        int len = words[0].size();        multimap<string, int> mp;        for(int i = 0; i < words.size(); ++i) mp.insert(pair<string, int>(words[i], i));        for(int i = 0; i < s.size() - num * len + 1; ++i){            string str = s.substr(i, num * len);            int k[num];            memset(k, 0, sizeof(k));            for(int j = 0; j < str.size(); j += len){                string tmp = str.substr(j, len);                if(mp.find(tmp) != mp.end()){                    pair <multimap<string,int>::iterator, multimap<string,int>::iterator> ret;                    ret = mp.equal_range(tmp);                    bool ok = false;                    for (multimap<string,int>::iterator it=ret.first; it!=ret.second; ++it){                        if(k[it->second] == 0){                            if(j == str.size() - len) vec.push_back(i);                            k[it->second] = 1;                            ok = true;                            break;                        }                    }                    if(!ok) break;                }                else break;            }        }        return vec;    }};

后来，改用unordered_map，使用1个映射记录words每个单词的数量，再使用一个映射记录字符串里各单词出现次数，如果出现次数大于对应单词数量，则说明不匹配。

class Solution {public:    vector<int> findSubstring(string s, vector<string>& words) {        vector<int> vec;        if(words.empty()) return vec;        int num = words.size();        int len = words[0].size();        unordered_map<string, int> mp;        for(string word : words)            mp[word]++;        for(int i = 0; i < s.size() - num * len + 1; ++i){            string str = s.substr(i, num * len);            unordered_map<string, int> mp2;            for(int j = 0; j < str.size(); j += len){                string word = str.substr(j, len);                if(mp.find(word) != mp.end()){                    mp2[word]++;                    if(mp2[word] > mp[word]) break;                }                else break;                if(j == str.size() - len) vec.push_back(i);            }        }        return vec;    }};