元音的子序列

Question

我正在练习面试并在网站上遇到这个问题:

字符串的神奇子序列S是按顺序S包含所有五个元音的子序列.找到字符串最大的魔法子序列的长度S.

例如,如果S = aeeiooua,然后aeiou和aeeioou是神奇的子序列,但aeio并aeeioua不是.

我是动态编程的初学者,我发现很难想出一个递归的公式.

Answer 1

我用一种迭代的方法而不是递归的方法。我开始构建类似于LIS（最长子序列）的解决方案，然后将其优化到O（n）。

#include<iostream>
#include<string>
#include<vector>
using namespace std;

string vowel = "aeiou";

int vpos(char c)
{
    for (int i = 0; i < 5; ++i)
        if (c == vowel[i])
            return i;
    return -1;
}

int magical(string s)
{
    int l = s.length();
    int previndex[5] = {-1, -1, -1, -1, -1};    // for each vowel
    vector<int> len (l, 0);
    int i = 0, maxlen = 0;

    // finding first 'a'
    while (s[i] != 'a')
    {
        ++i;
        if (i == l)
            return 0;
    }

    previndex[0] = i;       //prev index of 'a'
    len[i] = 1;

    for ( ++i; i < l; ++i)
    {
        if (vpos(s[i]) >= 0)    // a vowel
        {
            /* Need to append to longest subsequence on its left, only for this vowel (for any vowels) and 
             * its previous vowel (if it is not 'a')
                This important observation makes it O(n) -- differnet from typical LIS
            */
            if (previndex[vpos(s[i])] >= 0)
                len[i] = 1+len[previndex[vpos(s[i])]];

            previndex[vpos(s[i])] = i;

            if (s[i] != 'a')
            {
                if (previndex[vpos(s[i])-1] >= 0)
                    len[i] = max(len[i], 1+len[previndex[vpos(s[i])-1]]);
            }

            maxlen = max(maxlen, len[i]);
        }
    }
    return maxlen;
}

int main()
{
    string s = "aaejkioou";
    cout << magical(s);
    return 0;
}

Answer 2

#include <iostream>
#include<string>
#include<cstring>

using namespace std;
unsigned int getcount(string a, unsigned int l,unsigned int r );
int main()
{    
    std::string a("aaaaaeeeeaaaaiiioooeeeeuuuuuuiiiiiaaaaaaoo"
                 "oooeeeeiiioooouuuu");
    //std::string a("aaaaaeeeeaaaaiiioooeeeeuuuuuuiiiiiaaaaaaoooooeeeeiiioooo"); 
   //std::string a("aaaaaeeeeaaaaiiioooeeeeiiiiiaaaaaaoooooeeeeiiioooo"); //sol0
  //std::string a{"aeiou"};
  unsigned int len = a.length();
  unsigned int i=0,cnt =0,countmax =0;
  bool newstring = true;
  while(i<len)
  {
      if(a.at(i) == 'a' && newstring == true) 
      {
          newstring = false;
          cnt = getcount(a,i,len);
          if(cnt > countmax) 
          {
             countmax = cnt;
             cnt = 0;
          }
        } 
        else if(a.at(i)!='a')
        {
            newstring = true;
        }
        i++;
    }
    cout<<countmax;
    return 0;
}

unsigned int getcount(string a, unsigned int l,unsigned int r )
{
    std::string b("aeiou");
    unsigned int seq=0,cnt =0;
    unsigned int current =l;
    bool compstr = false;
    while(current<r)
    {
        if(a.at(current) == b.at(seq)) 
        {
            cnt++;
        }
        else if((seq <= (b.size()-2)) && (a.at(current) == b.at(seq+1)))
        {
            seq++; 
            cnt++;
            if (seq == 4) 
                compstr =true;
        }
        current++;
    }
    if (compstr == true) 
        return cnt;
   return 0;
}