Leetcode#30 Substring with concatenation of all Words

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Look at the words in L as a whole, which is similar to Minimun window string, which uses a sliding window to search.

So, in turn, enumerate all the start position I of the s, starting from I to search. Of course, it is not necessary to enumerate all i,i equal to the word length in L-1. For example, l in the word length of 3, then when the enumeration over i=0,1,2, no longer try to i=3, because the results have been covered by the i=0 situation.

Use the left and right two pointers (Left,right) to point to the left-to border of the window, moving the pointer as follows:

1. Expand the window. You might want to set the word "w" to the right and observe W:

If W is not in L, reset the window so that the window crosses W, i.e. right++, left = right

Otherwise, it may be possible to set the number of times that the W appears in the window is window[w],w in L TRAITS[W].

If WINDOW[W] < traits[w], stating that this word is not enough, then expand the window, move right to a unit, do not forget window[w]++

If window[w] = = Traits[w], indicating that the current window may constitute a set of solutions, then stop the expansion window and skip to step 3rd (check).

If WINDOW[W] > Traits[w], indicating that W appears too many times, can not be placed in the window, then stop the expansion window, jump to the 2nd step (shrinking window), the purpose is to remove the extra W window.

2. Shrink the window. The 1th step is to ensure that the words encountered at this time must appear in L.

Come here to explain that there must be a lot of words, you may assume that is W, then constantly right to move the left contraction window, each window[*left]--, until window[w] = = Traits[w], end contraction

3. Check if it is a set of solutions

This step is simple, see if the window length is equal to the length of L and can, because the first two steps to ensure that any word w must appear in L, and Window[w] <= traits[w]. If it is a solution, join the result set.

Then return to the 1th step and continue to the next cycle until the window moves to the s right edge.

Code:

1vector<int> findsubstring (stringS, vector<string> &L) {2vector<int>Res;3map<string,int>traits;4 5   if(L.empty () | | S.empty ())returnRes;6 7    for(auto s:l)8traits[s]++;9   intLen = l[0].length ();Ten  One    for(inti =0; i < Len; i++) { Amap<string,int>window; -  -     intL =i; the     intR =i; -      while(R <s.length ()) { -        while(R <s.length ()) { -         stringWord =S.substr (R, Len); +R + =Len; -         if(Traits.find (word) = =Traits.end ()) { + window.clear (); AL =R; at         } -         Else { -window[word]++; -           if(Window[word] >=Traits[word]) -              Break; -         } in       } -        while(L <r) { to         stringHead =s.substr (L, Len); +         stringTail = S.SUBSTR (R-Len, Len); -         if(Window[tail] = =Traits[tail]) the            Break; *L + =Len; $window[head]--;Panax Notoginseng       } -       if(R-l = = Len *l.size ()) { the Res.push_back (l); +       } A     } the   } +  -   returnRes; $}

Leetcode#30 Substring with concatenation of all Words