30. Substring with Concatenation of All Words

description:

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.
Note:

Example:

Example 1:

Input:
  s = "barfoothefoobarman",
  words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.
Example 2:

Input:
  s = "wordgoodgoodgoodbestword",
  words = ["word","good","best","word"]
Output: []

answer:

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        if (s.empty() || words.empty()) return {};
        vector<int> res;
        int n = words.size(), len = words[0].size();
        unordered_map<string, int> wordCnt;
        for (auto &word : words) ++wordCnt[word];
        //hashtabel统计词和对应的词频，上边这两句背一下，很有意思
        for (int i = 0; i <= (int)s.size() - n * len; ++i) {
        //这里大佬说一定要先把长度转成整数形式
            unordered_map<string, int> strCnt;
            int j = 0;
            for (j = 0; j < n ;j++){
                string t = s.substr(i + j * len, len);
                //活用c++的函数丫
                if (!wordCnt.count(t)) break;
                ++strCnt[t];
                if (strCnt[t] > wordCnt[t]) break;
            }
            if (j == n) res.push_back(i);
        }
        return res;
    }
};

relative point get√：

• string.substr

unordered_map<x,x> res;
for (auto &ele : eles) ++elecount[ele]

hint :

思路很简单，就是固定原字符串，然后拿单词表里的词去比，没有就过，多了就过。属于哈希表的活学活用吧。