java - Find largest sequence within an arraylist -

some background

last week did problem in textbook told me generate 20 random numbers , put brackets around successive numbers equal consider following program outputs

697342(33)(666)(44)69(66)1(88) 

what need do

the next problem longest sequence of these words , put brackets around them. if have

1122345(6666) 

basically need put brackets around 4 6's , since occur often. i've finished other problems in chapter studying ( arrays , arraylists), can't seem figure 1 out.

here solution have made putting brackets around successive numbers:

class seq {     private arraylist<integer> nums;     private random randnum;     public seq()     {         nums = new arraylist<integer>();         randnum = new random();     }     public void fillarrlist()     {         (int = 0 ; < 20 ; i++)         {             int thisrandnum = randnum.nextint(9)+1;             nums.add(thisrandnum);         }     }      public string tostring() {         stringbuilder result = new stringbuilder();         boolean inrun = false;         (int = 0; < nums.size(); i++) {             if (i < nums.size() - 1 && nums.get(i).equals(nums.get(i + 1))) {                 if (!inrun) {                     result.append("(");                 }                 result.append(nums.get(i));                 inrun = true;              } else {                 result.append(nums.get(i));                 if (inrun) {                     result.append(")");                 }                 inrun = false;              }         }         return result.tostring();     } } 

my thoughts

iterate through whole list. make count variable, keeps track of how many numbers successive of each other. i.e 22 have count of 2. 444 count of 3 next make oldcount, compares current count oldcount. want keep going if our new count greater oldcount

after need way starting index of largest count variable, end.

is way of thinking correct? because i'm having trouble updating oldcount , count variable while comparing them, since there values change. i'm not looking code, rather valuable hints.

my count resetting this

int startindex, endindex = 0;         int count = 0;         int oldcount = 0;              for(int = 0 ; < nums.size(); i++)             {                 if(nums.get(i) == nums.get(i+1) && count >= oldcount)                 {                     count++;                 }                 oldcount = count;             } 

only after walking elements know longest subsequence.

11222333333444555 11222(333333)444555 

hence after loop can insert both brackets.

so have maintain local optimum: start index plus length or last index of optimum. , every sequence start index of current sequence.

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as asked:

the optimal state (sequence) , current state 2 things. 1 cannot in advance current state final optimal state.

public string tostring() {     // begin "best" solution empty sequence.     int startbest = 0; // starting index     int lengthbest = 0; // length of sequence      // determine sequences:     int startcurrent = 0; // starting index of current/last sequence     (int = 0; < nums.size(); i++) {         // can add current num current sequence?         if (i == startcurrent || nums.get(i).equals(nums.get(i - 1)))) {             // can extend current sequence i:             int lengthcurrent = - startcurrent + 1;             if (lengthcurrent > lengthbest) { // current length better?                 // new optimum:                 startbest = startcurrent;                 lengthbest = lengthcurrent;             }         } else {             // different num, start here.             // had real sequence (i != 0), no need             // checking new optimum length 1.             startcurrent = i;         }     }     // found best solution.     // create result:     stringbuilder result = new stringbuilder();     (int = 0; < nums.size(); i++) {         result.append(nums.get(i));     }     // insert right ')' first index changes 1 after inserting '('.     result.insert(startbest + lengthbest, ")");     result.insert(startbest, "(");     return result.tostring(); } 

the first problem how find end of sequence, , set correct start of sequence.

the problem original algorithm there handled 1 sequence (one subsequence start).