One:
Hastset determines whether it is duplicated based on hashcode, and the data will not be repeated
Java code
/** List order not maintained **/ ashSet h = new HashSet (arlList);
arlList.clear();
arlList.addAll(h); Then do not add
Java code
/** List order maintained **/ public static void removeDuplicateWithOrder(ArrayList arlList) { Set set = new HashSet(); List newList = new ArrayList(); for (Iterator iter = arlList.iterator(); iter.hasNext(); ) { Object element = iter.next(); if (set.add(element)) newList.add(element); } arlList.clear(); arlList.addAll(newList); }
The following comes from the Internet:
Method 1: Loop element deletion
// 删除ArrayList中重复元素 public static void removeDuplicate(List list) { for ( int i = 0 ; i < list.size() - 1 ; i ++ ) { for ( int j = list.size() - 1 ; j > i; j -- ) { if (list.get(j).equals(list.get(i))) { list.remove(j); } } } System.out.println(list); }
Method 2: Eliminate through HashSet
// 删除ArrayList中重复元素 public static void removeDuplicate(List list) { HashSet h = new HashSet(list); list.clear(); list.addAll(h); System.out.println(list); }
Method 3: Delete duplicate elements in ArrayList and maintain order
// 删除ArrayList中重复元素,保持顺序 public static void removeDuplicateWithOrder(List list) { Set set = new HashSet(); List newList = new ArrayList(); for (Iterator iter = list.iterator(); iter.hasNext();) { Object element = iter.next(); if (set.add(element)) newList.add(element); } list.clear(); list.addAll(newList); System.out.println( " remove duplicate " + list); } 自己使用: 删除 “0.0”的值 List<List<String>> list1 = (List<List<String>>) map.get("商品入库表"); //表1 入库详细表 //删除list中 数量为 0值 for (Iterator<List<String>> item = list1.iterator(); item.hasNext(); ) { List<String> it = item.next(); System.out.print(it); if (it.get(4).equals("0.0")) { item.remove(); } }
Link address: http://iteye.blog.163.com/blog/static/186308096201302565345510/