我正在练习面试，并在网站上遇到了这个问题：
字符串
S的神奇子序列是S的子序列 按顺序包含所有五个元音。找到字符串S的最大魔法子序列的长度。例如，如果
S = aeeiooua，则aeiou和aeeioou是神奇的子序列 但aeio和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。
答案 0 :(得分：4)
我用迭代方法而不是递归方法做到了。我开始构建类似于LIS（最长增加子序列）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;
}
答案 1 :(得分：2)
O（输入字符串长度）运行时 导入java.util。*;
Item答案 2 :(得分：0)
#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;
}
答案 3 :(得分：0)
你可以在这里使用递归方法（这应该适用于字符串长度upto max int（可以使用容易记忆）
iTextSharp.text.Image logo = iTextSharp.text.Image.GetInstance("/test.png");
PdfPCell logocell = new PdfPCell(logo,true); // **PdfPCell(Image,Boolean Fit)**
}
答案 4 :(得分：0)
这是您的问题的python代码。 它是否使用非递归方法。
从我的存储库中挑选：MagicVowels Madaditya
############################################################################
#
# Magical Subsequence of Vowels
# by Aditya Kothari (https://github.com/Madaditya/magivvowels)
#
#
############################################################################
import sys
import re
usage = '''
Error : Invalid no of arguments passed.
Usage :
python magicv.py string_to_check
eg: python magicv.py aaeeiiooadasduu
'''
def checkMagicVowel(input_string):
#The Answer Variable
counter = 0
#Check if all vowels exist
if ('a' in input_string) and ('e' in input_string) and ('i' in input_string) and ('o' in input_string) and ('u' in input_string):
vowel_string = 'aeiou'
input_string_voweslOnly = ''
#Keeping only vowels in the string i.e user string MINUS NON vowels
for letter in input_string:
if letter in 'aeiou':
input_string_voweslOnly = input_string_voweslOnly + letter
magic = ''
index_on_vowel_string = 0
current_vowel = vowel_string[index_on_vowel_string]
#i is index on the current Character besing tested
i = 0
for current_char in input_string_voweslOnly:
if current_char == current_vowel:
counter = counter + 1
magic = magic + current_char
if (i < len(input_string_voweslOnly)-1):
next_char = input_string_voweslOnly[i+1]
if(index_on_vowel_string != 4):
next_vowel = vowel_string[index_on_vowel_string+1]
else:
next_vowel = vowel_string[index_on_vowel_string]
#next character should be the next new vowel only to count++
if (current_char != next_char and next_char == next_vowel):
if(index_on_vowel_string != 4):
index_on_vowel_string = index_on_vowel_string + 1
current_vowel = vowel_string[index_on_vowel_string]
i = i + 1
#Uncomment next line to print the all magic sequences
#print magic
'''
#Regex Method
#magic = re.match('[a]+[e]+[i]+[o]+[u]+',input_string_voweslOnly)
magic = re.match('[aeiou]+',input_string_voweslOnly)
if magic is not None:
##print magic.group()
##print len(magic.group())
else:
##print(0)
'''
else:
counter = 0
return counter
if __name__ == "__main__":
#checking arguments passed
if(len(sys.argv) != 2):
print usage
sys.exit(0)
input_string = sys.argv[1].lower()
#get all possible substrings
all_a_indices = [i for i, ltr in enumerate(input_string) if ltr == 'a']
substrings = []
for item in all_a_indices:
substrings.append(input_string[item:])
#print substrings
#Pass each substring and find the longest magic one
answer = []
for each in substrings:
answer.append(checkMagicVowel(each))
print max(answer)
答案 5 :(得分：0)
int func( char *p)
{
char *temp = p;
char ae[] = {'a','e','i','o','u'};
int size = strlen(p), i = 0;
int chari = 0, count_aeiou=0;
for (i=0;i<=size; i++){
if (temp[i] == ae[chari]) {
count_aeiou++;
}
else if ( temp[i] == ae[chari+1]) {
count_aeiou++;
chari++;
}
}
if (chari == 4 ) {
printf ("Final count : %d ", count_aeiou);
} else {
count_aeiou = 0;
}
return count_aeiou;
}
根据hackerrank挑战重新运行VOWELS计数的解决方案。
答案 6 :(得分：0)
int findsubwithcontinuousvowel(string str){
int curr=0;
int start=0,len=0,maxlen=0,i=0;
for(i=0;i<str.size();i++){
if(str[i]=='u' && (current[curr]=='u' || (curr+1<5 && current[curr+1]=='u'))){
//len++;
maxlen=max(len+1,maxlen);
}
if(str[i]==current[curr]){
len++;
}
else if(curr+1<5 && str[i]==current[curr+1]){
len++;
curr++;
}
else{
len=0;
curr=0;
if(str[i]=='a'){
len=1;
}
}
}
return maxlen;
}
答案 7 :(得分：0)
检查元音在isInSequence中是否可用，并在processor上处理结果。
public class one {
private char[] chars = {'a','e','i','o','u'};
private int a = 0;
private boolean isInSequence(char c){
// check if char is repeating
if (c == chars[a]){
return true;
}
// if vowels are in sequence and just passed by 'a' and so on...
if (c == 'e' && a == 0){
a++;
return true;
}
if (c == 'i' && a == 1){
a++;
return true;
}
if (c == 'o' && a == 2){
a++;
return true;
}
if (c == 'u' && a == 3){
a++;
return true;
}
return false;
}
private char[] processor(char[] arr){
int length = arr.length-1;
int start = 0;
// In case if all chars are vowels, keeping length == arr
char array[] = new char[length];
for (char a : arr){
if (isInSequence(a)){
array[start] = a;
start++;
}
}
return array;
}
public static void main(String args[]){
char[] arr = {'m','a','e','l','x','o','i','o','u','a'};
one o = new one();
System.out.print(o.processor(arr));
}
}
答案 8 :(得分：0)
#include <bits/stdc++.h>
#define ios ios::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL);
#define ll unsigned long long
using namespace std;
int main() {
// your code goes here
ios
string s;
cin>>s;
int n=s.length();
int dp[n+1][5]={0};
for(int i=1;i<=n;i++)
{
if(s[i-1]=='a')
{
dp[i][0]=1+dp[i-1][0];
dp[i][1]=dp[i-1][1];
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='e')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][0]>0)
{dp[i][1]=1+max(dp[i-1][1],dp[i-1][0]);}
else
dp[i-1][1]=0;
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='i')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][1]>0)
{dp[i][2]=1+max(dp[i-1][1],dp[i-1][2]);}
else
dp[i-1][2]=0;
dp[i][1]=dp[i-1][1];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='o')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][2]>0)
{dp[i][3]=1+max(dp[i-1][3],dp[i-1][2]);}
else
dp[i-1][3]=0;
dp[i][2]=dp[i-1][2];
dp[i][1]=dp[i-1][1];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='u')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][3]>0)
{dp[i][4]=1+max(dp[i-1][4],dp[i-1][3]);}
else
dp[i-1][4]=0;
dp[i][1]=dp[i-1][1];
dp[i][3]=dp[i-1][3];
dp[i][2]=dp[i-1][2];
}
else
{
dp[i][0]=dp[i-1][0];
dp[i][1]=dp[i-1][1];
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
}
cout<<dp[n][4];
return 0;
}