如何从列表中删除重复项？ [英] How to remove duplicates from a list?

问题描述

我想从列表中删除重复的邮件，但是我所做的不起作用：

 列表< Customer> listCustomer = new ArrayList< Customer>（）; 
（客户：tmpListCustomer）
 {
 if（！listCustomer.contains（customer））
 {
 listCustomer.add 
} 
} 
  

解决方案

该代码不起作用，您可能没有在 Customer 类上适当地实现 equals（Object） p>

大概有一些密钥（让我们称为 customerId ）唯一标识一个客户;例如

  class Customer {
 private String customerId; 
 ... 
  

看起来像这样：

  public boolean equals（Object obj）{
 if obj == this）{
 return true; 
} 
 if（！（obj instanceof Customer））{
 return false; 
} 
 Customer other =（Customer）obj; 
 return this.customerId.equals（other.customerId）; 
} 
  

为了完整起见， c $ c> hashCode ，这样两个相等的 Customer 对象将返回相同的哈希值。 equals 的上述定义的匹配 hashCode 将是：

  public int hashCode（）{
 return customerId.hashCode（）; 
} 
  

这也是值得注意的，这不是一个有效的方法来删除重复if列表很大。 （对于具有N个客户的列表，您将需要在最差情况下执行 N *（N-1）/ 2 比较;即当没有重复时）。一个更有效的解决方案，你应该使用像 HashSet 来做重复检查。

I want to remove duplicates from a list but what I am doing is not working:

List<Customer> listCustomer = new ArrayList<Customer>();    
for (Customer customer: tmpListCustomer)
{
  if (!listCustomer.contains(customer)) 
  {
    listCustomer.add(customer);
  }
 }

解决方案

If that code doesn't work, you probably have not implemented equals(Object) on the Customer class appropriately.

Presumably there is some key (let us call it customerId) that uniquely identifies a customer; e.g.

class Customer {
    private String customerId;
    ...

An appropriate definition of equals(Object) would look like this:

    public boolean equals(Object obj) {
        if (obj == this) {
            return true;
        }
        if (!(obj instanceof Customer)) {
            return false;
        }
        Customer other = (Customer) obj;
        return this.customerId.equals(other.customerId);
    }

For completeness, you should also implement hashCode so that two Customer objects that are equal will return the same hash value. A matching hashCode for the above definition of equals would be:

    public int hashCode() {
        return customerId.hashCode();
    }

It is also worth noting that this is not an efficient way to remove duplicates if the list is large. (For a list with N customers, you will need to perform N*(N-1)/2 comparisons in the worst case; i.e. when there are no duplicates.) For a more efficient solution you should use something like a HashSet to do the duplicate checking.

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