Find the intersection of N lists contained in SearchResults

listintersection.java

if (SearchResults.size() == 1) {
    DATA = SearchResults.get(0);
} else {
    /* Find the smallest list */
    int baseitem = 0;
    for (int i = 1; i < SearchResults.size(); i++) {
        if (SearchResults.get(i).size() < SearchResults.get(baseitem).size()) {
            baseitem = i;
        }
    }
    
    /* Initialize DATA to be the maximum size it could need to be */
    DATA = new ArrayList<type>(SearchResults.get(baseitem).size());
    
    /* Now take the intersection of those lists, looping through one list and comparing */
    for (DataType Record : SearchResults.get(baseitem)) {
        Boolean matched = true;
        for (int i = 0; i < SearchResults.size(); i++) {
            if (i == baseitem) { 
                break;
            }
            if (!SearchResults.get(i).contains(Record)) {
                matched = false;
                break;
            }
        }
        if (matched) {
            DATA.add(Record);
        }
    }
    
    DATA.trimToSize();
}

Would be faster if we saved position in each secondary list as items in the primary were found, as in sorted results the new results cannot be found before the old ones.