


How to determine whether there is an intersection between two collections in java
Nov 20, 2020 pm 03:13 PMBackground:
The front-end passes the list collection, and the back-end fields are also in the form of (1,2,3,4). Without using sql, how to check whether the set passed by the front end is in the set of back-end fields?
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Code:
public static boolean judgeIntersection(List<String> list1,List<String> list2){ boolean flag = false; // 使用retainAll會改變list1的值,所以寫一個替代 List<String> origin = new ArrayList<>(); origin.addAll(list1); origin.retainAll(list2); // 有交集 if(origin.size()>0){ flag = true; } return flag; }
boolean flag = origin.retainAll(Collection> ; c)
If there is data in origin that is not in set C, return false. No true is returned
At the same time, the origin set will change, and only the same data of the two sets will be retained. In other words, the origin set has deleted data, and false will be returned
How to judge whether there is Intersection?
1. Two sets, none of them are the same, origin is empty, and the return value is false
2. Two sets, by chance, the data in origin are all in c, origin No change, the return value is true
These two special cases lead to the fact that it is impossible to judge the intersection simply by relying on the return value true or false. Therefore, depending on the number of origin collections, >0, there is an intersection
##retainAll(Collection> c) source code
public boolean retainAll(Collection<?> c) { // 判斷c集合是否為空 Objects.requireNonNull(c); return batchRemove(c, true); } private boolean batchRemove(Collection<?> c, boolean complement) { // 得到調(diào)用該函數(shù)的集合。因為是引用類型,所以修改了都會有變化 final Object[] elementData = this.elementData; // w:記錄交集的數(shù)據(jù)都放到elementData前面,w是其位置分界線 int r = 0, w = 0; // 返回值 用于判斷elementData有沒有被修改 boolean modified = false; try { // 循環(huán)elementData集合,判斷其中元素是否在c集合中 for (; r < size; r++) // 若在集合中,則w自增,并將該值放到elementData[w]中,即是交集的數(shù)據(jù)都放到集合的前面 if (c.contains(elementData[r]) == complement) elementData[w++] = elementData[r]; } finally { // Preserve behavioral compatibility with AbstractCollection, // even if c.contains() throws. // 正常情況下,經(jīng)過上面的循環(huán),r==size。為防止出現(xiàn)循環(huán)異常,將由于異常導致的r到size是交集的數(shù)據(jù)但并沒有放到對應w的位置的數(shù)據(jù),都放到對應w之后的位置上 if (r != size) { System.arraycopy(elementData, r, elementData, w, size - r); w += size - r; } // 若elementData中有數(shù)據(jù)不在c集合中,就清理掉w位置之后的數(shù)據(jù),便于垃圾回收 if (w != size) { // clear to let GC do its work for (int i = w; i < size; i++) elementData[i] = null; // 記錄elementData集合被增刪的次數(shù),這里是刪除 modCount += size - w; // 賦值最新的size size = w; // elementData集合被清理,modified為true modified = true; } } return modified; }Related recommendations:
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