List of combination

Assuming that there are N list, list which has M integers . How kind these N list seamless stitching together . For example aList have 123456, bList have 45678, cList 789 . This time , aList with cList on seamless stitching up into dList: 123456789.
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cycle to repeat , but the efficiency is not the same in different ways

List<Integer> l1 = new ArrayList<Integer>();
  		List<Integer> l2 = new ArrayList<Integer>();
  		for (int i = 1; i < 7; i++) {
  			l1.add(i);
  		}
  		for (int i = 4; i < 9; i++) {
  			l2.add(i);
  		}
  		l1.addAll(l2);
  
  		Set set = new HashSet();
  		List newList = new ArrayList();
  		for (Iterator iter = l1.iterator(); iter.hasNext();) {
  			Object element = iter.next();
  			if (set.add(element))
  				newList.add(element);
  		}
  		for (Object o : newList) {
  			System.out.print(o);
  		}
  		System.out.println("\n---");
  		List<Integer> listInt = new ArrayList<Integer>(
  				new LinkedHashSet<Integer>(l1));
  		for (Integer o : listInt) {
  			System.out.print(o);
  		}

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list sorting relatively combination
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1, HashSet, the List which in turn put a number to it they will go heavy.
2, if you have to synthesize a List, then on the basis of a through ArrayList (Collection c) constructor new one List
3, or create an empty ArrayList, when the other side of the inside contains comparisons are included with
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determine whether the elements under both meets plus a principle would be all right . The first tail +1 is equal to the second one first . Intermediate elements would not have considered .
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123 and 789 count seamlessly ? Or , it must be completely linked , orderly integer it?

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you list all of what you put into a string a good amount . index is not any stitching , and vice versa on the fight .
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public static void main(String args[]) throws Exception{
  		List list1=new ArrayList();
  		List list2=new ArrayList();
  		
  		list1.add("1");
  		list1.add("2");
  		list2.add("2");
  		List<String> newChar=new ArrayList();
  		concat(newChar,list1,list2);
  		for(int i=0;i<newChar.size();i++){
  			System.out.println(newChar.get(i));
  		}
  	}
  	
  	static List concat(List<String> newChar,List<String> list1,List<String> list2){
  		String[] char1=list1.toArray(new String[0]);
  		String[] char2=list2.toArray(new String[0]);
  	
  		for(int i=0;i<char1.length;i++){
  			newChar.add(char1[i]);
  		}
  		for(int j=0;j<char2.length;j++){
  			if(!newChar.contains(char2[j])) newChar.add(char2[j]);
  			else {
  				System.out.println("有重复元素，不能拼接");
  				newChar.removeAll(newChar);
  				break;
  			}
  		}
  		return newChar;
  	}

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premise list already is uninterrupted integer combinations, try the first one that +1 is equal to the second end of the first , was then on for the value to the list , and not on the continue;

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completely linked , orderly   .
ah , ideas probably do two operations: one new to the list to be seamless with the original list ( refer to 5L ideas ) ; 2 passed a judgment, but also new to the list . as ordered. This long traverse once , kept equal to +1 .
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I express is not very clear, I mean, stitching together the two list .
example aList have 123456, bList have 45678, cList 789 . This time , aList with cList on seamless stitching up into dList: 123456789. And aList and bList is not seamless stitching together because there are duplicate 456 between them. bList and cList can not be spliced ​​together , they exist duplicate 78.


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completely linked , orderly