集合合并

用于时间表的合并
时间段的合并工作.

public class StarEnds  {
	LinkedList<StarEnd> qeuqeu = new LinkedList<StarEnd>();
	public StarEnds add(StarEnd other) {
		Iterator<StarEnd> it = qeuqeu.iterator();
		while(it.hasNext()){
			StarEnd temp = it.next();
			if(StarEnd.isMix(temp, other)){
				other.inputOther(temp);
				it.remove();
			}
		}
		qeuqeu.add(other);
		return this;
	}	
	public Iterator<StarEnd> iterator(){
		return qeuqeu.iterator();
	}
	 

	public static void main(String[] args) {
		StarEnds ses = new StarEnds();
		int[] arr = {1,2,4,5};
		for(int i : arr){
			StarEnd temp = new StarEnd();
			temp.star=i;
			temp.end =i+1;
			ses.add(temp);
		}
		System.out.println(ses.qeuqeu);
	}


}
 class StarEnd{
	int star=0; 
	int end=0;
	@Override
	public String toString() {
		return "{"+star+":"+end+"}";
	}
	public void inputOther(StarEnd other){
		    List<Integer> l = Arrays.asList(star,end,other.star,other.end);
			star =  Collections.min(l);
			end =   Collections.max(l);
			other.star = 0;
			other.end = 0;
	
	}
	public static boolean isMix(StarEnd one, StarEnd other) {
		if(one==null||other==null){
			return false;
		}
		return one.inStarEnd(other.star)||one.inStarEnd(other.end) ;
	}
	public boolean inStarEnd(int in){
		return (in >= star && in<= end);
	}
}

评论

1 楼 weiqiang.yang 2010-11-12  

这里的start应该是小于end的吧

 List<Integer> l = Arrays.asList(star,end,other.star,other.end);  
            star =  Collections.min(l);  
            end =   Collections.max(l);  

这里可以直接比较star 和 other.star得到最小，end 和other.end得到最大，也许可以减少几次比较:D

如果start不一定小于end，那么isMix方法判断貌似就不全面了