leetcode-30-Substring with Concatenation of All Words
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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.
For example, given:
s: “barfoothefoobarman”
words: [“foo”, “bar”]
You should return the indices: [0,9].
题目大意:
s是str,words是包含若干个str的list。找到s中的所有位置,从这些位置开始往后可以包括words中所有str。包含的顺序无关,但是要次数相同。比如在words中有出现3次的字符串要同样在该位置后一段距离出现3次。
思路:
遍历words。用字典哈希表保存words中每一个元素出现的次数。
遍历s。到达每一个位置后,每次往后截取一个word长度的元素保存在temp中,判断temp是否在map里。维护一个临时tmpmap保存每个s的位置往后出现的words中元素的次数。维护一个break条件flag。
当某次截取的temp不在map里,flag为false,break。
当某word在tmpmap中次数大于map中次数,flag置false,break。
开始想的是每次遍历s的某一位置都copy一个words,然后截取一个word,在words中就在words中将该元素删除。知道某个word不在words中或者words为空时退出。思路简洁没问题。
结果超时。words元素多了以后时间复杂度巨大。要遍历s长度次words。
用字典只需要遍历一次words。
python代码:
class Solution(object): def findSubstring(self, s, words): """ :type s: str :type words: List[str] :rtype: List[int] """ lens=len(s) lenw=len(words) len0=len(words[0]) map={} res=[] for i in words: if i in map:map[i]+=1 else:map[i]=1 for i in range(lens-lenw*len0+1): tmpmap={} for j in range(lenw): flag=True k=i+j*len0 temp=s[k:k+len0] if temp in map: num=0 if temp in tmpmap:num=tmpmap[temp] if map[temp]<num+1:flag=False;break tmpmap[temp]=num+1 else: flag=False break if flag:res.append(i) return res
AC。700+ms。
c++代码:
也要700+ms。思路一样。
不知道结果里面100以内怎么做到的。。
class Solution {public: vector<int> findSubstring(string S, vector<string> &L) { map<string, int> words; map<string, int> curWords; vector<int> ret; int slen = S.length(); if (!slen || L.empty()) return ret; int llen = L.size(), wlen = L[0].length(); // record the current words map for (auto &i : L) ++words[i]; // check the [llen * wlen] substring for (int i = 0; i + llen * wlen <= slen; i++) { curWords.clear(); int j = 0; // check the words for (j = 0; j < llen; j++) { string tmp = S.substr(i + j * wlen, wlen); if (words.find(tmp) == words.end()) break; ++curWords[tmp]; if (curWords[tmp] > words[tmp]) break; } if (j == llen) ret.push_back(i); } return ret; }};
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