[size=x-large]请问：怎样对List集合中的数据进行排序啊？？ 请给与详解！！！
谢谢[/size]
[size=large][color=red]要对List排序,你要对List里装的这种类型的类实现排序接口(Comparable).
举个例子:
下面这个例子是对List进行排序.使用Collections.sort(List list);方法,此方法会调用MyObject的compareTo方法.所以在MyObject类定义里要实现compareTo方法.
[code="java"]public class ListSort {
/**
* main()
* 2010-4-2,下午09:25:57
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
List<MyObject> lists = new ArrayList<MyObject>();
MyObject obj1 = new MyObject("d");
MyObject obj2 = new MyObject("a");
lists.add(obj1);
lists.add(obj2);
Collections.sort(lists);
for (MyObject myObject : lists) {
System.out.println(myObject.getContent());
}
}
}
class MyObject implements Comparable{
private String content;
public MyObject(String _content) {
this.content = _content;
}
public String getContent() {
return content;
}
public void setContent(String content) {
this.content = content;
}
public int compareTo(MyObject obj) {
// TODO Auto-generated method stub
if (null == obj) return 1;
else {
return this.content.compareTo(obj.content);
}
}
}[/code]
下面是Collections.sort方法
[code="java"]public static > void sort(List list) {
Object[] a = list.toArray();
Arrays.sort(a);
ListIterator i = list.listIterator();
for (int j=0; j<a.length; j++) {
i.next();
i.set((T)a[j]);
}
}[/code]
实际上我们的MyObject类的方法compareTo是在Arrays.sort()中被调用的...
请看..
[code="java"]public static void sort(Object[] a) {
Object[] aux = (Object[])a.clone();
mergeSort(aux, a, 0, a.length, 0);
}[/code]
[code="java"] private static void mergeSort(Object[] src,
Object[] dest,
int low,
int high,
int off) {
int length = high - low;
// Insertion sort on smallest arrays
if (length < INSERTIONSORT_THRESHOLD) {
for (int i=low; i<high; i++)
for (int j=i; j>low &&
((Comparable) dest[j-1]).compareTo(dest[j])>0; j--)
swap(dest, j, j-1);
return;
}
// Recursively sort halves of dest into src
int destLow = low;
int destHigh = high;
low += off;
high += off;
int mid = (low + high) >> 1;
mergeSort(dest, src, low, mid, -off);
mergeSort(dest, src, mid, high, -off);
// If list is already sorted, just copy from src to dest. This is an
// optimization that results in faster sorts for nearly ordered lists.
if (((Comparable)src[mid-1]).compareTo(src[mid]) <= 0) {
System.arraycopy(src, low, dest, destLow, length);
return;
}
// Merge sorted halves (now in src) into dest
for(int i = destLow, p = low, q = mid; i < destHigh; i++) {
if (q >= high || p < mid && ((Comparable)src[p]).compareTo(src[q])<=0)
dest[i] = src[p++];
else
dest[i] = src[q++];
}
}[/code][/color][/size]
当然,这是一种实现排序的办法.还有一种是实现Comparator,实现这个接口,然后使用
Collections.sort(List list,Comparator c);这个方法来排序..
原理差不多..
希望对你有帮助..
冒泡可以啊！
for (int i = 1; i < list.size; i++) {
String temp = (String)list.get(i);
for (int j = 0; j < i; j++) {
if(compareTo(list.get(j).)>0){
String temp = sourceAry[i];
sourceAry[i]=sourceAry[j];
sourceAry[j]=temp;
}
}
}
实现一个自定义Comparator，然后使用Collections.sort方法
冒泡可以！我这里是对数组做排序，你改成list就可以了。
String[] sourceAry = {"1","6","4","8","3","9"};
for (int i = 1; i < sourceAry.length; i++) {
for (int j = 0; j < i; j++) {
if(sourceAry[i].compareTo(sourceAry[j])>0){
String temp = sourceAry[i];
sourceAry[i]=sourceAry[j];
sourceAry[j]=temp;
}
}
}
java.util.Collections类的sort方法也可以实现。
public static class Comparator1 implements Comparator {
/**
* 功能：对单位进行比较
*
* @return 1, o1在o2之后
* -1, o1在o2之前
*/
public int compare(Object o1, Object o2) {
String[] ary1 = (String[]) o1;
String[] ary2 = (String[]) o2;
if (ary1[3].equals(ary2[3])) { // 1. 是同级别
if (ary1[8].equals("") && ary2[8].equals("")) {
return 0;
} else if (ary1[8].equals("")) {
return 1;
} else if (ary2[8].equals("")) {
return -1;
} else {
return Integer.parseInt(ary1[8])-Integer.parseInt(ary2[8]);
}
} else { // 2. 不是同级别
if (ary1[3].equals("") && ary2[3].equals("")) {
return 0;
} else if (ary1[3].equals("")) {
return 1;
} else if (ary2[3].equals("")) {
return -1;
} else {
return ary1[3].compareTo(ary2[3]);
}
}
}
}
Collections.sort(dwCodeList, new Comparator1());
Comparator cmp = new Comparator(){
@Override
public int compare(String o1, String o2) {
int len1 = o1.length();
int len2 = o2.length();
int len = Math.min(len1, len2);
for(int i=0;i if(o2.charAt(i) > o1.charAt(i)){
return 1;
}
else if(o2.charAt(i) < o1.charAt(i)){
return -1;
}
}
//if之前的都相等,看谁更长一些
if(len2 > len1){
return 1;
}
else if(len2 < len1){
return -1;
}
else{
return 0;
}
}
};
你要再不会自己思考，那就真的太懒了。