高效比较两个list中不同的元素

╰+攻爆jí腚メ 2022-12-03 05:14 284阅读 0赞

第四种方法，将其中一个list转为map,并通过containsKey()方法比较，用在单元测试，读取dbf文件到redis队列，测试redis和dbf的是否相同

package com.test;
    
    import java.util.ArrayList;
    import java.util.HashMap;
    import java.util.List;
    
    public class Test {
        public static void main(String[] args) {
            HashMap<String, String> map = new HashMap<>();
    
            List<String> list1=new ArrayList();
            list.add("one");
    
            List<String> list2=new ArrayList();
            list.add("one");
    
            for (String str1:list1) {
                map.put(str1,"1");
            }
    
            for (String str2:list2) {
                map.containsKey(str2);
            }
        }
    }

###  参考博客： ###

package com.syl.test;
     
    import java.util.*;
     
    /**
     * 获取两个List的不同元素（假设List自身不存在重复元素）
     * Created by syl on 2017/12/26 0026.
     */
    public class TestCompareList {
        public static void main(String[] args) {
            List<String> list1 = new ArrayList<String>();
            List<String> list2 = new ArrayList<String>();
            for (int i = 0; i < 30000; i++) {
                list1.add("test" + i);
            }
            for (int i = 0; i < 80000; i++) {
                list2.add("test" + i * 2);
            }
     
            getDiffrent1(list1, list2);
            getDiffrent2(list1, list2);
            getDiffrent3(list1, list2);
            getDiffrent4(list1, list2);
        }
     
        // 方法1，两层遍历查找，遍历次数为list1.size()*list2.size()，有点蠢
        private static List<String> getDiffrent1(List<String> list1, List<String> list2) {
            // diff 存放不同的元素
            List<String> diff = new ArrayList<String>();
            // 开始查找的时间，用于计时
            long start = System.currentTimeMillis();
            for (String str : list1) {
                if (!list2.contains(str)) {
                    diff.add(str);
                }
            }
            // 计时
            System.out.println("方法1 耗时：" + (System.currentTimeMillis() - start) + " 毫秒");
            return diff;
        }
     
        // 方法2，两层遍历查找，用retainAll()方法查找，也很蠢，方法底层依旧是两层遍历
        private static List<String> getDiffrent2(List<String> list1, List<String> list2) {
            long start = System.currentTimeMillis();
            list1.retainAll(list2);// 返回值是boolean
            System.out.println("方法2 耗时：" + (System.currentTimeMillis() - start) + " 毫秒");
            return list1;
        }
     
        // 方法3，用Map存放List1和List2的元素作为key，value为其在List1和List2中出现的次数
        // 出现次数为1的即为不同元素，查找次数为list1.size() + list2.size()，较方法1和2，是极大简化
        private static List<String> getDiffrent3(List<String> list1, List<String> list2) {
            List<String> diff = new ArrayList<String>();
            long start = System.currentTimeMillis();
            Map<String, Integer> map = new HashMap<String, Integer>(list1.size() + list2.size());
            // 将List1元素放入Map，计数1
            for (String string : list1) {
                map.put(string, 1);
            }
            // 遍历List2，在Map中查找List2的元素，找到则计数+1；未找到则放入map，计数1
            for (String string : list2) {
                Integer count = map.get(string);
                if (count != null) {
                    map.put(string, ++count);// 此处可优化，减少put次数，即为方法4
                    continue;
                }
                map.put(string, 1);
            }
            for (Map.Entry<String, Integer> entry : map.entrySet()) {
                if (entry.getValue() == 1) {
                    diff.add(entry.getKey());
                }
            }
            System.out.println("方法3 耗时：" + (System.currentTimeMillis() - start) + " 毫秒");
            return diff;
        }
     
        // 优化方法3，减少put次数
        private static List<String> getDiffrent4(List<String> list1, List<String> list2) {
            List<String> diff = new ArrayList<String>();
            long start = System.currentTimeMillis();
            Map<String, Integer> map = new HashMap<String, Integer>(list1.size() + list2.size());
            List<String> maxList = list1;
            List<String> minList = list2;
            if (list2.size() > list1.size()) {
                maxList = list2;
                minList = list1;
            }
            for (String string : maxList) {
                map.put(string, 1);
            }
            for (String string : minList) {
                Integer count = map.get(string);
                if (count != null) {
                    map.put(string, ++count);
                    continue;
                }
                map.put(string, 1);
            }
            for (Map.Entry<String, Integer> entry : map.entrySet()) {
                if (entry.getValue() == 1) {
                    diff.add(entry.getKey());
                }
            }
            System.out.println("方法4 耗时：" + (System.currentTimeMillis() - start) + " 毫秒");
            return diff;
     
        }
     
    }

转载地址：[https://blog.csdn.net/sinat\_32133675/article/details/79386072][https_blog.csdn.net_sinat_32133675_article_details_79386072]

[https_blog.csdn.net_sinat_32133675_article_details_79386072]: https://blog.csdn.net/sinat_32133675/article/details/79386072