获取两个字符串List的相同元素的两种方法。

红太狼 2024-04-18 14:46 101阅读 0赞

时间复杂度：O( nlog(n) )  
输出结果集为Collection子类：如图  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L0J1dHRlcmZseV9yZXN0aW5n_size_16_color_FFFFFF_t_70]

/**
         * 获取两个字符串List的相同元素
         *
         * @param listOne 字符串列表1
         * @param listTwo 字符串列表2
         * @param result  用于装结果集的容器
         * @return 返回两个字符串List的相同元素
         */
        private static <T extends  Collection<String>> T getAllSameElementOnTwoList(List<String> listOne, List<String> listTwo, T result) {
            if (listOne.isEmpty() || listTwo.isEmpty()) {
                return result;
            }
            Collections.sort(listOne);
            Collections.sort(listTwo);
    
            int k = 0;
            int j = 0;
            while (k < listOne.size() && j < listTwo.size()) {
                if (listOne.get(k).compareTo(listTwo.get(j)) == 0) {
                    if (listOne.get(k).equals(listTwo.get(j))) {
                        result.add(listOne.get(k));
                        k++;
                        j++;
                    }
                } else if (listOne.get(k).compareTo(listTwo.get(j)) < 0) {
                    k++;
                } else {
                    j++;
                }
            }
            return result;
        }

示例：容器为HashSet，对相同的元素进行去重。

public static void main(String[] args) {
            List<String> list1 = new ArrayList<String>();
            List<String> list2 = new ArrayList<String>();
            list1.add("ab");
            list1.add("abc");
            list1.add("ab");
            list1.add("abcd");
            list1.add("a");
    
            list2.add("d");
            list2.add("abc");
            list2.add("abcd");
            //获取当前时间戳
            long last = System.currentTimeMillis();
            for (String value : Objects.requireNonNull(getAllSameElementOnTwoList(list1, list2, new HashSet<>()))) {
                System.out.print(value + " ");
            }
            System.out.println();
            long now = System.currentTimeMillis();
            System.out.println(now - last);
        }
    
        //abc abcd
    }

方法二，使用List接口的retainAll来取交集

> /\*\*  
> \* Retains only the elements in this list that are contained in the  
> \* specified collection (optional operation). In other words, removes  
> \* from this list all of its elements that are not contained in the  
> \* specified collection.  
> \*  
> \* @param c collection containing elements to be retained in this list  
> \* @return true if this list changed as a result of the call  
> \* @throws UnsupportedOperationException if the retainAll operation  
> \* is not supported by this list  
> \* @throws ClassCastException if the class of an element of this list  
> \* is incompatible with the specified collection  
> \* (optional)  
> \* @throws NullPointerException if this list contains a null element and the  
> \* specified collection does not permit null elements  
> \* (optional),  
> \* or if the specified collection is null  
> \* @see \#remove(Object)  
> \* @see \#contains(Object)  
> \*/  
> boolean retainAll(Collection<?> c);

使用案例：

如果需要改变原list的可以用方法二，不想改原list结构的话可以使用方法一。

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L0J1dHRlcmZseV9yZXN0aW5n_size_16_color_FFFFFF_t_70]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/18/964d27887e1f466bbc09b3b4626887d7.png