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标题: ArrayList存储自定义对象,并且去除重复元素的一个问题 [打印本页]

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作者: 王亚飞    时间: 2013-7-3 11:40
标题: ArrayList存储自定义对象,并且去除重复元素的一个问题

1. //ArrayList打印自定义对象,并且去除重复元素
   
2. 
   
3. import java.util.*;
   
4. class Person
   
5. {
   
6. private String name;
   
7. private int age;
   
8. Person(String name,int age){
   
9. this.name = name;
   
10. this.age = age;
    
11. }
    
12. public String getName(){
    
13. return name;
    
14. }
    
15. public int getAge(){
    
16. return age;
    
17. }
    
18. public boolean equals(Object obj){
    
19. if(!(obj instanceof Person))
    
20. return false;
    
21. Person p = (Person)obj;
    
22. return this.name.equals(p.name)&&this.age == p.age;
    
23. }
    
24. }
    
25. 
    
26. class UserDefind {
    
27. public static void main(String[] args) {
    
28. 
    
29. ArrayList al = new ArrayList();
    
30. al.add(new Person("张三",15));
    
31. al.add(new Person("张si",56));
    
32. al.add(new Person("张si",56));
    
33. al.add(new Person("张wu",87));
    
34. al.add(new Person("张liu",38));
    
35. al.add(new Person("张liu",38));
    
36. al.add(new Person("张qi",45));
    
37. 
    
38. al = method(al);//.............................就是行,报错不兼容的类型
    
39. 
    
40. Iterator it = al.iterator();
    
41. while(it.hasNext()){
    
42. Person p = (Person)it.next();
    
43. System.out.println(p.getName()+"--"+p.getAge());
    
44. }
    
45. 
    
46. 
    
47. }
    
48. public static List method(ArrayList al){
    
49. ArrayList newal = new ArrayList();
    
50. 
    
51. Iterator lit = al.iterator();
    
52. while(lit.hasNext()){
    
53. Object obj = lit.next();
    
54. 
    
55. if(!newal.contains(obj))
    
56. newal.add(obj);
    
57. 
    
58. }
    
59. return newal;
    
60. }
    
61. }
    

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38行不兼容的类型.....怎么会出现这个错误,修改方法

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作者: j816326    时间: 2013-7-3 11:47

1. //ArrayList打印自定义对象,并且去除重复元素
   
2. 
   
3. 
   
4. import java.util.*;
   
5. 
   
6. class Person
   
7. 
   
8. {
   
9. 
   
10. private String name;
    
11. 
    
12. private int age;
    
13. 
    
14. Person(String name,int age){
    
15. 
    
16. this.name = name;
    
17. 
    
18. this.age = age;
    
19. 
    
20. }
    
21. 
    
22. public String getName(){
    
23. 
    
24. return name;
    
25. 
    
26. }
    
27. 
    
28. public int getAge(){
    
29. 
    
30. return age;
    
31. 
    
32. }
    
33. 
    
34. public boolean equals(Object obj){
    
35. 
    
36. if(!(obj instanceof Person))
    
37. 
    
38. return false;
    
39. 
    
40. Person p = (Person)obj;
    
41. 
    
42. return this.name.equals(p.name)&&this.age == p.age;
    
43. 
    
44. }
    
45. 
    
46. }
    
47. 
    
48. 
    
49. class UserDefind {
    
50. 
    
51. public static void main(String[] args) {
    
52. 
    
53. 
    
54. ArrayList al = new ArrayList();
    
55. 
    
56. al.add(new Person("张三",15));
    
57. 
    
58. al.add(new Person("张si",56));
    
59. 
    
60. al.add(new Person("张si",56));
    
61. 
    
62. al.add(new Person("张wu",87));
    
63. 
    
64. al.add(new Person("张liu",38));
    
65. 
    
66. al.add(new Person("张liu",38));
    
67. 
    
68. al.add(new Person("张qi",45));
    
69. 
    
70. 
    
71. al = method(al);//.............................就是行,报错不兼容的类型
    
72. 
    
73. 
    
74. Iterator it = al.iterator();
    
75. 
    
76. while(it.hasNext()){
    
77. 
    
78. Person p = (Person)it.next();
    
79. 
    
80. System.out.println(p.getName()+"--"+p.getAge());
    
81. 
    
82. }
    
83. 
    
84. 
    
85. 
    
86. }
    
87. 
    
88. public static List method(ArrayList al){
    
89. 
    
90. ArrayList newal = new ArrayList();
    
91. 
    
92. 
    
93. Iterator lit = al.iterator();
    
94. 
    
95. while(lit.hasNext()){
    
96. 
    
97. Object obj = lit.next();
    
98. 
    
99. 
    
100. if(!newal.contains(obj))
     
101. 
     
102. newal.add(obj);
     
103. 
     
104. 
     
105. }
     
106. 
     
107. return newal;
     
108. 
     
109. }
     
110. }
     

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method 的方法的返回值是List 但是你接收的却是ArrayList 所以出错了。该下返回值， 或者接收的类型改变下

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作者: 王亚飞    时间: 2013-7-3 11:49

j816326 发表于 2013-7-3 11:47
method 的方法的返回值是List 但是你接收的却是ArrayList 所以出错了。该下返回值， 或者接收的类型改变下 ...

果然.....多谢,还有个疑问,List不是ArrayList的父类嘛,这样也不行?

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作者: 王靖远    时间: 2013-7-3 11:50
public static List method(ArrayList al){ 的、你把这里的返回值类型改成ArrayList试试啊

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作者: 王亚飞    时间: 2013-7-3 11:53

王靖远 发表于 2013-7-3 11:50
public static List method(ArrayList al){ 的、你把这里的返回值类型改成ArrayList试试啊 ...

恩恩,确实是这个问题

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作者: 小石头39910    时间: 2013-7-3 12:30
因为你用Arraylist创建的对象a1所以在调用函数的时候返回值必须是ArrayList类型

1. <p>//ArrayList打印自定义对象,并且去除重复元素</p><p>import java.util.*;
   
2. class Person
   
3. {
   
4. private String name;
   
5. private int age;
   
6. Person(String name,int age){
   
7. this.name = name;
   
8. this.age = age;
   
9. }
   
10. public String getName(){
    
11. return name;
    
12. }
    
13. public int getAge(){
    
14. return age;
    
15. }
    
16. public boolean equals(Object obj){
    
17. if(!(obj instanceof Person))
    
18. return false;
    
19. Person p = (Person)obj;
    
20. return this.name.equals(p.name)&&this.age == p.age;
    
21. }
    
22. }</p><p>class UserDefind {
    
23. public static void main(String[] args) {</p><p>ArrayList al = new ArrayList();
    
24. al.add(new Person("张三",15));
    
25. al.add(new Person("张si",56));
    
26. al.add(new Person("张si",56));
    
27. al.add(new Person("张wu",87));
    
28. al.add(new Person("张liu",38));
    
29. al.add(new Person("张liu",38));
    
30. al.add(new Person("张qi",45));</p><p>al = method(al);//.............................就是行,报错不兼容的类型</p><p>Iterator it = al.iterator();
    
31. while(it.hasNext()){
    
32. Person p = (Person)it.next();
    
33. System.out.println(p.getName()+"--"+p.getAge());
    
34. }</p><p>
    
35. }
    
36. public static ArrayList method(ArrayList al){
    
37. ArrayList newal = new ArrayList();</p><p>Iterator lit = al.iterator();
    
38. while(lit.hasNext()){
    
39. Object obj = lit.next();</p><p>if(!newal.contains(obj))
    
40. newal.add(obj);</p><p>}
    
41. return newal;
    
42. }
    
43. }
    
44. </p><p> </p>

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