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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.
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 = "wordgoodstudentgoodword",
words = ["word","student"]
Output: []
算法实现:
- class Solution {
- public:
- vector<int> findSubstring(string s, vector<string>& words) {
- vector<int> res;
- if(s.empty()) return res;
- if(words.empty()) return res;
- unordered_map<string, int> m;
- unordered_map<string, int> mt;
- int word_nums = words.size();
- int word_size = words[0].length();
- for(int i = 0; i < words.size(); i ++)m[words[i]] ++;
- int j;
- for(int i = 0; i < s.length(); i ++){
- mt.clear();
- for(j = 0; j < word_nums; j ++)
- {
- string word = s.substr(i + j*word_size, word_size);
- if(m.count(word))
- {
- if(mt[word] < m[word])mt[word] ++;
- else break;
- }
- else break;
- }
- if(j == word_nums) res.push_back(i);
- }
- return res;
- }
- };
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