如何优雅查找两个存储自定义对象的ArrayList中各自独有的字符串

核心优化思路 #

• 提前抽取两个列表的目标字符串属性存入HashSet，利用Set O(1)时间复杂度的查找能力，将原实现O(n*m)的时间复杂度降低为O(n+m)，数据量越大性能提升越明显
• 逻辑清晰简洁，无需重复编写反向比对的遍历代码

优化后代码 #

import java.util.ArrayList;
import java.util.Set;
import java.util.stream.Collectors;

public class CompareTwoArrays {

    ArrayList<MyCustomObject> firstArrayListOfObjects = new ArrayList<>();
    ArrayList<MyCustomObject> secondArrayListOfObjects = new ArrayList<>();

    public void superSpecificExampleMethod() {
        firstArrayListOfObjects.add(new MyCustomObject(1, 1, "apple"));
        firstArrayListOfObjects.add(new MyCustomObject(1, 1, "orange"));
        firstArrayListOfObjects.add(new MyCustomObject(1, 1, "banana"));
        firstArrayListOfObjects.add(new MyCustomObject(1, 1, "grape"));
        secondArrayListOfObjects.add(new MyCustomObject(1, 1, "apple"));
        secondArrayListOfObjects.add(new MyCustomObject(1, 1, "pineapple"));
        secondArrayListOfObjects.add(new MyCustomObject(1, 1, "orange"));
        secondArrayListOfObjects.add(new MyCustomObject(1, 1, "banana"));

        // 提前提取两个列表的字符串属性生成Set，用于快速查找
        Set<String> firstStrSet = firstArrayListOfObjects.stream()
                .map(MyCustomObject::getString)
                .collect(Collectors.toSet());
        Set<String> secondStrSet = secondArrayListOfObjects.stream()
                .map(MyCustomObject::getString)
                .collect(Collectors.toSet());

        // 查找仅存在于第一个列表的字符串
        firstArrayListOfObjects.stream()
                .map(MyCustomObject::getString)
                .filter(str -> !secondStrSet.contains(str))
                .forEach(str -> System.out.println(str + " doesn't exist in second arrayList"));

        // 查找仅存在于第二个列表的字符串
        secondArrayListOfObjects.stream()
                .map(MyCustomObject::getString)
                .filter(str -> !firstStrSet.contains(str))
                .forEach(str -> System.out.println(str + " doesn't exist in first arrayList"));
    }
}

class MyCustomObject {

    private int randomIntOne;
    private int randomIntTwo;
    private String string;

    public MyCustomObject(int randomIntOne, int randomIntTwo, String string) {
        this.randomIntOne = randomIntOne;
        this.randomIntTwo = randomIntTwo;
        this.string = string;
    }

    public String getString() {
        return string;
    }
}

场景适配 #

如果只需要获取无重复的差异字符串集合，可以直接用Set求差集，代码更精简：

// 仅在第一个列表存在的字符串（无重复）
Set<String> onlyInFirst = new HashSet<>(firstStrSet);
onlyInFirst.removeAll(secondStrSet);
// 仅在第二个列表存在的字符串（无重复）
Set<String> onlyInSecond = new HashSet<>(secondStrSet);
onlyInSecond.removeAll(firstStrSet);

内容的提问来源于stack exchange，提问作者Apple