java - Really need two passes in this algorithm? -

suppose want perform action on each word if sentence contains >= 4 numbers. example,

• this string contains 10, 23, 30, 50, 60.
• this string contains 10, , contains 23.

i need take particular action on first sentence above, since contains more 4 numbers. in algorithm above came with, needs 2 passes of string list. there more efficient algorithm task? instance, 1 pass only? thank you.

void process (list<string> sentence) {  boolean has4abovenum = false; if(containsmorethan4numbers(sentence)) {     has4abovenum = true; }  for(string word : sentence) {    if(has4abovenum) {       dosomething(word);      } }    }  boolean containsmorethan4numbers(list<string> sentence) { int numcount = 0; for(string word : sentence) {   if(numcount>4){      return true;   }    if(isnumber(word)) {       numcount++;   }    } return false; }  

you can on 1 path parsing string, , creating in 1 pass on string list<string> contains numbers in string, each element in list distinct number found in string.

now, have check list.size() - , make sure in desired range.

list<string> getnumbers(list<string> sentence) {    list<string> res = new arraylist<>();    for(string word : sentence)  {      if(isnumber(word)) res.add(word);    }    return res; }