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).
下面这段代码只能通过三分之二的测试用例,因为a concatenation of each word in wordsexactly once,我能说题目意思给的不明确么?
网上还有应用“滑动窗口”的时间复杂度为O(n)算法,后面再分析!
1 class Solution { 2 public: 3 vector<int> findSubstring(string s, vector<string>& words) { 4 vector<int> res; 5 int wordNum=words.size(),wordLen=words[0].size(); 6 if(s.size()<wordLen*wordNum) 7 return res; 8 map<string,int> dic,curWord; 9 for(int i=0;i<wordNum;i++) 10 { 11 dic[words[i]]++; 12 } 13 for(int i=0;i<s.size()-wordLen*wordNum;i++) 14 { 15 curWord.clear(); 16 int count=0; 17 for(int j=0;j<wordNum;j++) 18 { 19 string word=s.substr(i+j*wordLen,wordLen); 20 if(dic.find(word)==dic.end()) 21 break; 22 curWord[word]++; 23 count++; 24 if(curWord[word]>dic[word]) 25 break; 26 } 27 if(count==wordNum) 28 res.push_back(i); 29 } 30 return res; 31 } 32 };