【Java】两个 List 遍历匹配数据的优化处理

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首先，举个例子：例如有一个List<Student>，另外一个是List<Husband>，你现在要筛选出来自同一个家庭的丈夫和妻子，组成一个家庭的数据全集（一个家庭包含丈夫、妻子），这样就是吧数据赋值给另外一个List<Family>（这个例子可能不恰当，欢迎评论给出更好的例子）;

说明：代码采用java8的Stream和Lambda处理List，如果对Stream和Lambda不太了解的，先查阅另外一篇文章：http://blog.csdn.net/qi_lin7/article/details/52958123

好吧，大家还是结合代码来看吧

Wife.java

import java.io.Serializable;

/**
* Author:
* Date: 2016/11/23.
* Time: 18:08
*/
public class Wife implements Serializable {
private int id;
/**妻子名字**/
private String wifeName;
/**家庭ID**/
private String familyId;

public Wife() {

}

public Wife(int id, String wifeName, String familyId) {
this.id = id;
this.wifeName = wifeName;
this.familyId = familyId;
}

public int getId() {
return id;
}

public void setId(int id) {
this.id = id;
}

public String getWifeName() {
return wifeName;
}

public void setWifeName(String wifeName) {
this.wifeName = wifeName;
}

public String getFamilyId() {
return familyId;
}

public void setFamilyId(String familyId) {
this.familyId = familyId;
}

@Override
public String toString() {
return "";
}
}

Husband.java

import java.io.Serializable;

/**
* Author:
* Date: 2016/11/23.
* Time: 18:08
*/
public class Husband implements Serializable {
private int id;
/**丈夫名字**/
private String husbandName;
/**家庭ID**/
private String familyId;

public int getId() {
return id;
}

public void setId(int id) {
this.id = id;
}

public Husband(){

}

public Husband(int id, String wifeName, String familyId) {
this.id = id;
this.husbandName = wifeName;
this.familyId = familyId;
}

public String getHusbandName() {
return husbandName;
}

public void setHusbandName(String husbandName) {
this.husbandName = husbandName;
}

public String getFamilyId() {
return familyId;
}

public void setFamilyId(String familyId) {
this.familyId = familyId;
}

@Override
public String toString() {
return "";
}
}

Family.java

import java.io.Serializable;

/**
* Author:
* Date: 2016/11/23.
* Time: 18:08
*/
public class Family implements Serializable {
/**妻子名字**/
private String wifeName;
/**丈夫名字**/
private String husbandName;
/**家庭ID**/
private String familyId;

public String getWifeName() {
return wifeName;
}

public void setWifeName(String wifeName) {
this.wifeName = wifeName;
}

public String getHusbandName() {
return husbandName;
}

public void setHusbandName(String husbandName) {
this.husbandName = husbandName;
}

public String getFamilyId() {
return familyId;
}

public void setFamilyId(String familyId) {
this.familyId = familyId;
}

@Override
public String toString() {
return "";
}
}

TestFamily.java

import com.xxx.bean.Family;
import com.xxx.bean.Husband;
import com.xxx.bean.Wife;
import com.google.common.collect.Lists;
import org.junit.Test;

import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

/**
* Author:
* Date: 2016/11/23.
* Time: 18:17
*/
public class TestFamily {
@Test
public void familyTest() {
List<Wife> wife = Lists.newArrayList();
wife.add(new Wife(0, "我是张三的妻子", "0001"));
wife.add(new Wife(0, "我是李四的妻子", "0002"));
wife.add(new Wife(0, "我是王五的妻子", "0003"));

List<Husband> husband = Lists.newArrayList();
husband.add(new Husband(0, "我是张三", "0001"));
husband.add(new Husband(0, "我是李四", "0002"));
husband.add(new Husband(0, "我是王五", "0003"));

List<Family> families = Lists.newArrayList();
// 将list转为Map,这里key一定要为唯一值
Map<String, Wife> wifeMap = wife.stream().collect(
Collectors.toMap(w -> w.getFamilyId(),
w -> w));
// 匹配家庭
families = husband.stream().map(h -> {
return toFamily(wifeMap.get(h.getFamilyId()), h);
}).collect(Collectors.toList());

families.stream().forEach(family -> {
System.out.println("家庭ID：" + family.getFamilyId() + ",丈夫：" + family.getHusbandName() + ",妻子：" + family.getWifeName());
});
}

private Family toFamily(Wife wife, Husband husband) {
Family family = new Family();
family.setFamilyId(wife.getFamilyId());
family.setHusbandName(husband.getHusbandName());
family.setWifeName(wife.getWifeName());
return family;
}
}

最后输出结果：

家庭ID：0001,丈夫：我是张三,妻子：我是张三的妻子

家庭ID：0002,丈夫：我是李四,妻子：我是李四的妻子

家庭ID：0003,丈夫：我是王五,妻子：我是王五的妻子

这种处理办法，可以将处理此类问题的循环次数由N*N，变为2*N

如果不理解或者有更好的解决办法处理此类问题的