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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 wordsexactly 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).
思路: 判断一个值是否包含在一个数组中，首先应该想到将这个数组中的元素放入HashTable，否则每次查找都需要O(n)的时间复杂度。
时间复杂度：O(n*size)，其中n为s的长度，size是指数组words包含多少个元素。当words元素不多的时候，我们可以说时间复杂度是线性的O(n)
class Solution { public: vector<int> findSubstring(string s, vector<string>& words) { size = words.size(); sLen = s.length(); wLen = words[0].length(); wordsLen = wLen * size; for(i = 0; i < size; i++){ word_counter[words[i]]++; } i = 0; while(i+wordsLen<=sLen){ for(j = 0; j < size; j++){ cmpStr = s.substr(i+j*wLen, wLen); if(word_counter.find(cmpStr)==word_counter.end()){ break; } counting[cmpStr]++; if(counting[cmpStr]>word_counter[cmpStr]){ break; } } if(j==size){ ret.push_back(i); } counting.clear(); i++; } return ret; } private: string cmpStr; vector<int> ret; map<string,int> word_counter; map<string,int> counting; int size; //number of words int sLen; int wLen; int wordsLen; int i; int j; };
30. Substring with Concatenation of All Words (HashTable)
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原文地址：http://www.cnblogs.com/qionglouyuyu/p/4823330.html
