一、场景：

二、结论：

1. 四种方法耗时

三、代码：

---

一、场景：

• 求差集 List1 - Lsit2

二、结论：

1. 四种方法耗时

初始条件	方法名	方法思路	耗时
List1.size=319418 List2.size=284900	List..removeAll(Lsit2)		1036987ms
	removeAll_01	List.contains()	614859ms
	removeAll_02	运用Set	150ms	推荐
	removeAll_03	用Set.contains()再优化	112ms	推荐

 

三、代码：

package com.privatecloud.core.util.collections;import com.alibaba.fastjson2.JSON;
import com.privatecloud.core.util.file.FileIOUtil;
import com.privatecloud.core.util.file.FilesReadUtil;
import com.privatecloud.core.util.file.FilesUtil;
import lombok.extern.slf4j.Slf4j;import java.io.IOException;
import java.util.*;@Slf4j
public class ListUtils<T> {public List<T> removeAll_01(List<T> source, List<T> destination) {List<T> result = new LinkedList<T>();for (T t : source) {if (!destination.contains(t)) {result.add(t);}}return result;}/*** 2，运用Set可以去重这一特性。效率有明显提升** @param source* @param destination* @return*/public List<T> removeAll_02(List<T> source, List<T> destination) {List<T> result = new LinkedList<T>();Map<T, Integer> sourceMap = new HashMap<T, Integer>();for (T t : source) {if (sourceMap.containsKey(t)) { //原集合中的重复值sourceMap.put(t, sourceMap.get(t) + 1);} else {sourceMap.put(t, 1);}}Set<T> all = new HashSet<T>(destination);for (Map.Entry<T, Integer> entry : sourceMap.entrySet()) {T key = entry.getKey();Integer value = entry.getValue();if (all.add(key)) {for (int i = 0; i < value; i++) {result.add(key);}}}return result;}/*** 3，用Set.contains()再优化** @param source* @param destination* @return*/public List<T> removeAll_03(List<T> source, List<T> destination) {List<T> result = new LinkedList<T>();Set<T> destinationSet = new HashSet<T>(destination);for (T t : source) {if (!destinationSet.contains(t)) {result.add(t);}}return result;}}