[Classic] X (Long | Yamato | Large product) Y (sub-sequence | Sub-string)

Note: Sub-sequences, which can be discontinuous;

The following topics, as I see it, are difficult to rank from easy to difficult:

Max and sub-sequences (Maximum sum subsequence)

This problem is purely entertaining ... There should be no such problem. The scenario is to put the sequence into a positive number.

vector<int> msseq (vector<int> &num) {    vector<int> result;      for (int  i:num)          if 0 )            result.push_back (i);     return result;}

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Maximum increment substring (longest increasing substring)

The violence programme TC is O (n^2). A better scheme is, the greedy method , set i from 0 to n traversal, with Gmax for the global longest length, Len represents the length of the current increment substring. When the increment is encountered, the len++ is encountered, the GMAX is updated, and the Len is reset to 1. TC = O (n), SC = O (1)

stringLisstringstr) {    if(Str.empty ())returnstr; stringRecstr ="", Curstr =""; Curstr+ = str[0]; intGmax = int_min, Len =1;  for(inti =1; I < str.length (); i++) {        if(Str[i] > Str[i-1]) {len++; Curstr+=Str[i]; }        Else {            //assume that you only need to return one of the longest, and if you want to return all the vectors are stored            if(Gmax <Len) {Gmax=Len; Recstr=Curstr; } curstr=""; Len=1; }    }    returnrecstr;}

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Maximum increment subsequence (longest increasing subsequence)

The violence programme TC is O (2^n). If using the greedy method, the analysis know that according to the information contained in the first I can not cover the number of cases, so the decision of the number of i+1 cannot be done; a better scheme is, dynamic planning , with Opt[i] record to the longest and the longest increment subsequence containing the characters I have.

vector<int> Miseq (vector<int> &num) {Vector<int>result; if(Num.empty ())returnnum; Vector<int> Opt (num.size (),1), Record (Num.size (),-1);  for(inti =0; I < num.size (); i++) {         for(intj =0; J < I; J + +) {            if(Num[i] > Num[j] && opt[j] +1>Opt[i]) {Opt[i]= Opt[j] +1; Record[i]=J; }        }    }    intLast =-1, Gmax =0;  for(inti =0; I < num.size (); i++) {        if(Opt[i] >gmax) {Gmax=Opt[i]; Last=i; }    }     while(Last >=0) {result.push_back (num[last]); Last=Record[last];    } reverse (Result.begin (), Result.end ()); returnresult;}

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Maximum no repetition string

Longest matching substring

Edit Distance

Max and sub-sequences

Maximum product sub-sequence

Maximum Product substring

[Classic] X (Long | Yamato | Large product) Y (sub-sequence | Sub-string)