对于此作业,我将编写一个方法removeDuplicates,该方法将排序的字符串ArrayList 作为参数,并从列表中消除任何重复项。
例如,假设名为 list 的变量包含以下值:
{"be", "be", "is", "not", "or", "question", "that", "the", "to", "to"}
调用removeDuplicates(list)后,列表应存储以下值:
{"be", "is", "not", "or", "question", "that", "the", "to"}
我几乎已经把它记下来了,但由于某种原因,如果列表包含
["duplicate", "duplicate", "duplicate", "duplicate", "duplicate"]
它将删除除两个之外的所有内容,从而导致[重复,重复]而不是
这是我的代码:
private static void removeDuplicates(ArrayList<String> thing) {
for (int i = 0; i < thing.size(); i++) { // base word to compare to
String temp = thing.get(i);
for (int j = 0; j < thing.size(); j++) { // goes through list for match
String temp2 = thing.get(j);
if (temp.equalsIgnoreCase(temp2) && i != j) { // to prevent removal of own letter.
thing.remove(j);
}
}
}
}
最佳答案
问题是,即使找到重复项,您也会执行“j++”。一旦你做了“thing.remove(j);”它本质上是将所有内容向下移动一个索引值,因此您不必增加 j。
示例:
{ duplicate, duplicate, duplicate, duplicate, duplicate }
i iteration 1:
i=0 [dup, dup, dup, dup, dup]
j=0 [dup, dup, dup, dup, dup]
remove=1
j=1 [dup, dup, dup, dup]
remove=2
j=2 [dup, dup, dup]
i iteration 2:
i=1 [dup, dup, dup]
remove=0
j=0 [dup, dup]
j=1 [dup, dup]
[dup, dup]
i iteration 3 stops since 3>size of list.
关于java - 删除 ArrayList 中的重复项,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19261292/