为什么在 ArrayList 中写入冒泡排序时出错？
[英]Why i have error when write buble sort in ArrayList?
问题描述
I have ArrayList with some words.
我有带一些词的 ArrayList。 I need to sort words by way of repetition.我需要通过重复的方式对单词进行排序。 Thanks.谢谢。
ArrayList<String> words = new ArrayList<String>();
words.addAll(Arrays.asList("returns", "such", "containing", "created", "maintain", "such", "discipline", "specified", "these", "such", "which", "such", "larger",
"deceptively", "difficult", "returns", "element", "care", "returns", "collection"));
int i = 0, j = 0;
String sortTemp;
for (j = 0; j < words.size(); j++) {
for (i = 0; i < words.size()-j; i++) {
if (words.get(i).compareTo(words.get(i+1))>0) {
sortTemp = words.get(i);
words.set(i, words.get(i+1));
words.set(i+1, sortTemp);
}
}
}
for(i=0;i<words.size(); i++) {
Log.e("Sorted Array", words.get(i));
}
01-13 06:09:31.065 740-740/?
2 个解决方案
解决方案1
0 2016-01-13 12:41:25
There is no need to implement bubble-sort.不需要实现冒泡排序。 You can just use the sort method你可以只使用sort方法
ArrayList<String> words = new ArrayList<String>();
words.addAll(Arrays.asList("returns", "such", "containing", "created", "maintain", "such", "discipline", "specified", "these", "such", "which", "such", "larger",
"deceptively", "difficult", "returns", "element", "care", "returns", "collection"));
Collections.sort(words):
After this code fragment your ArrayList will be sorted在此代码片段之后，您的 ArrayList 将被排序
As for the error you are getting, you can clearly see that it is an IndexOutOfBounds Exception, which means the index of your arraylist elements is greater than it's size至于你得到的错误，你可以清楚地看到它是一个IndexOutOfBounds Exception，这意味着你的arraylist元素的索引大于它的大小
for(int i = 0; i < words.size(); i++) {
for(int j = 0; j < words.size()-i-1; j++) { // you make the size smaller here
if(words.get(j).compareTo(words.get(j+1)) > 0) {
//....
}
}
}
解决方案2
0
Try this,尝试这个，
ArrayList<String> words = new ArrayList<String>();
words.addAll(Arrays.asList("returns", "such", "containing", "created", "maintain", "such", "discipline", "specified", "these", "such", "which", "such", "larger",
"deceptively", "difficult", "returns", "element", "care", "returns", "collection"));
String tempString;
for(int i = 0; i < words.size(); i++) {
for(int j = 0; j < words.size() - 1 -i; j++) {
if(words.get(j).compareTo(words.get(j+1)) > 0) {
tempString = words.get(j);
words.set(j,words.get(j+1));
words.set(j+1,tempString);
}
}
}
for (int index = 0 ; index < words.size(); index ++){
Log.d("MainActivity","element @ "+index+" : "+words.get(index));
}
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