对List去重并保证添加顺序主要有三种方式

Myth丶恋晨 2021-10-30 04:28 295阅读 0赞

方式一，利用HashSet不能添加重复数据的特性 由于HashSet**不能保证添加顺序**，所以**只能作为判断条件**：

private static void removeDuplicate(List<String> list) {
        HashSet<String> set = new HashSet<String>(list.size());
        List<String> result = new ArrayList<String>(list.size());
        for (String str : list) {
            if (set.add(str)) {  //如果能成插入Set，说明这个元素之前没有出现过。即这个元素不是重复的
                result.add(str);
            }
        }
        list.clear();
        list.addAll(result);
    }

方式二，利用LinkedHashSet不能添加重复数据，并能**保证添加顺序不变**的特性 ：

private static void removeDuplicate(List<String> list) {
        LinkedHashSet<String> set = new LinkedHashSet<String>(list.size());
        set.addAll(list);
        list.clear();
        list.addAll(set);
    }

方式三，利用List的contains方法循环遍历：

private static void removeDuplicate(List<String> list) {
        List<String> result = new ArrayList<String>(list.size());
        for (String str : list) {
            if (!result.contains(str)) {
                result.add(str);
            }
        }
        list.clear();
        list.addAll(result);
    }

准备测试程序：

import java.util.ArrayList;
    import java.util.HashSet;
    import java.util.LinkedHashSet;
    import java.util.List;
    
    public class Test {
        private static void removeDuplicate1(List<String> list) {
            HashSet<String> set = new HashSet<String>(list.size());
            List<String> result = new ArrayList<String>(list.size());
            for (String str : list) {
                if (set.add(str)) {
                    result.add(str);
                }
            }
            list.clear();
            list.addAll(result);
        }
    
        private static void removeDuplicate2(List<String> list) {
            LinkedHashSet<String> set = new LinkedHashSet<String>(list.size());
            set.addAll(list);
            list.clear();
            list.addAll(set);
        }
    
        private static void removeDuplicate3(List<String> list) {
            List<String> result = new ArrayList<String>(list.size());
            for (String str : list) {
                if (!result.contains(str)) {
                    result.add(str);
                }
            }
            list.clear();
            list.addAll(result);
        }
    
    
        public static void main(String[] args) {
            final List<String> list = new ArrayList<String>();
            for (int i = 0; i < 1000; i++) {
                list.add("haha-" + i);
            }
    
            long time = System.currentTimeMillis();
            for (int i = 0; i < 10000; i++) {
                removeDuplicate1(list);
            }
    
            long time1 = System.currentTimeMillis();
            System.out.println("time1:"+(time1-time));
    
    
            for (int i = 0; i < 10000; i++) {
                removeDuplicate2(list);
            }
            long time2 = System.currentTimeMillis();
            System.out.println("time2:"+(time2-time1));
    
            for (int i = 0; i < 10000; i++) {
                removeDuplicate3(list);
            }
            long time3 = System.currentTimeMillis();
            System.out.println("time3:"+(time3-time2));
    
    
        }
    }

结果为：

> time1:563  
> time2:625  
> time3:16418

从上面可以看出，方法3的耗时非常久！

总结：从便捷以及效率上，方式二是最佳选择，具体原因可以参考HashMap的存储方式  
[http://www.cnblogs.com/chenssy/p/3521565.html][http_www.cnblogs.com_chenssy_p_3521565.html]

[http_www.cnblogs.com_chenssy_p_3521565.html]: http://www.cnblogs.com/chenssy/p/3521565.html