LeetCode:Substring with Concatenation of All Words （summarize）

题目链接

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

 

算法1

暴力解法，从字符串s的每个位置都判断一次（如果从当前位置开始的子串长度小于L中所有单词长度，不用判断），从当前位置开始的子串的前段部分能不能由集合L里面的单词拼接而成。

从某一个位置 i 判断时，依次判断单词s[i,i+2], s[i+3,i+5], s[i+6, i+8]…是否在集合中，如果单词在集合中，就从集合中删除该单词。

我们用一个hash map来保存单词，这样可以在O(1)时间内判断单词是否在集合中

算法的时间复杂度是O（n*(l*k)）n是字符串的长度，l是单词的个数，k是单词的长度

 

递归代码如下：

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class Solution { private:     int wordLen;       public:     vector<int> findSubstring(string S, vector<string> &L) {         unordered_map<string, int>wordTimes;         for(int i = 0; i < L.size(); i++)             if(wordTimes.count(L[i]) == 0)                 wordTimes.insert(make_pair(L[i], 1));             else wordTimes[L[i]]++;         wordLen = L[0].size();                   vector<int> res;         for(int i = 0; i <= (int)(S.size()-L.size()*wordLen); i++)             if(helper(S, i, wordTimes, L.size()))                 res.push_back(i);         return res;     }       //判断子串s[index...]的前段是否能由L中的单词组合而成     bool helper(string &s, const int index,         unordered_map<string, int>&wordTimes, const int wordNum)     {         if(wordNum == 0)return true;         string firstWord = s.substr(index, wordLen);         unordered_map<string, int>::iterator ite = wordTimes.find(firstWord);         if(ite != wordTimes.end() && ite->second > 0)         {             (ite->second)--;             bool res = helper(s, index+wordLen, wordTimes, wordNum-1);             (ite->second)++;//恢复hash map的状态             return res;         }         else return false;     } };

 

非递归代码如下：

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class Solution { private:     int wordLen;       public:     vector<int> findSubstring(string S, vector<string> &L) {         unordered_map<string, int>wordTimes;         for(int i = 0; i < L.size(); i++)             if(wordTimes.count(L[i]) == 0)                 wordTimes.insert(make_pair(L[i], 1));             else wordTimes[L[i]]++;         wordLen = L[0].size();                   vector<int> res;         for(int i = 0; i <= (int)(S.size()-L.size()*wordLen); i++)             if(helper(S, i, wordTimes, L.size()))                 res.push_back(i);         return res;     }           //判断子串s[index...]的前段是否能由L中的单词组合而成     bool helper(const string &s, int index,         unordered_map<string, int>wordTimes, int wordNum)     {         for(int i = index; wordNum != 0 && i <= (int)s.size()-wordLen; i+=wordLen)         {             string word = s.substr(i, wordLen);             unordered_map<string, int>::iterator ite = wordTimes.find(word);             if(ite != wordTimes.end() && ite->second > 0)                 {ite->second--; wordNum--;}             else return false;         }         if(wordNum == 0)return true;         else return false;     } };

 

OJ递归的时间小于非递归时间，因为非递归的helper函数中，hash map参数是传值的方式，每次调用都要拷贝一次hash map，递归代码中一直只存在一个hash map对象

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算法2

回想前面的题目：LeetCode:Longest Substring Without Repeating Characters 和 LeetCode:Minimum Window Substring ，都用了一种滑动窗口的方法。这一题也可以利用相同的思想。

比如s = “a1b2c3a1d4”L={“a1”,“b2”,“c3”，“d4”}

窗口最开始为空，

a1在L中，加入窗口 【a1】b2c3a1d4                            本文地址

b2在L中，加入窗口 【a1b2】c3a1d4

c3在L中，加入窗口 【a1b2c3】a1d4

a1在L中了，但是前面a1已经算了一次，此时只需要把窗口向右移动一个单词a1【b2c3a1】d4

d4在L中，加入窗口a1【b2c3a1d4】找到了一个匹配

如果把s改为“a1b2c3kka1d4”，那么在第四步中会碰到单词kk，kk不在L中，此时窗口起始位置移动到kk后面a1b2c3kk【a1d4

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class Solution { public:     vector<int> findSubstring(string S, vector<string> &L) {         unordered_map<string, int>wordTimes;//L中单词出现的次数         for(int i = 0; i < L.size(); i++)             if(wordTimes.count(L[i]) == 0)                 wordTimes.insert(make_pair(L[i], 1));             else wordTimes[L[i]]++;         int wordLen = L[0].size();                   vector<int> res;         for(int i = 0; i < wordLen; i++)         {//为了不遗漏从s的每一个位置开始的子串，第一层循环为单词的长度             unordered_map<string, int>wordTimes2;//当前窗口中单词出现的次数             int winStart = i, cnt = 0;//winStart为窗口起始位置,cnt为当前窗口中的单词数目             for(int winEnd = i; winEnd <= (int)S.size()-wordLen; winEnd+=wordLen)             {//窗口为[winStart,winEnd)                 string word = S.substr(winEnd, wordLen);                 if(wordTimes.find(word) != wordTimes.end())                 {                     if(wordTimes2.find(word) == wordTimes2.end())                         wordTimes2[word] = 1;                     else wordTimes2[word]++;                                           if(wordTimes2[word] <= wordTimes[word])                         cnt++;                     else                     {//当前的单词在L中，但是它已经在窗口中出现了相应的次数，不应该加入窗口                      //此时，应该把窗口起始位置想左移动到，该单词第一次出现的位置的下一个单词位置                         for(int k = winStart; ; k += wordLen)                         {                             string tmpstr = S.substr(k, wordLen);                             wordTimes2[tmpstr]--;                             if(tmpstr == word)                             {                                 winStart = k + wordLen;                                 break;                             }                             cnt--;                         }                     }                                           if(cnt == L.size())                         res.push_back(winStart);                 }                 else                 {//发现不在L中的单词                     winStart = winEnd + wordLen;                     wordTimes2.clear();                     cnt = 0;                 }             }         }         return res;     } };

算法时间复杂度为O（n*k)）n是字符串的长度，k是单词的长度

 

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