java 如何快速求一个字符串数组中出现最多次数的字符串?字符串数组存在大量重复.要最快速度.

这样就OK了，你试一下
class Test
{
public static void main(String[] args)
{
int max=1;
String[] strs={"a","b","f","a","d","c","e","d","a","d"};
for(int i=0;i {
int count=1;
for(int j=i+1;j {
if(strs[i].equals(strs[j]))
count++;
}
if(max max=count;
}
System.out.println("重复最多次数为："+max);
for(int i=0;i {
int count=1;
for(int j=i+1;j {
if(strs[i].equals(strs[j]))
count++;
}
if(count==max)
System.out.println("重复最多次("+max+")的字符串为："+strs[i]);
}
}
}

public static String[] arrayUnique(String[] a){
List list = new LinkedList();
for (int i = 0; i < a.length; i++) {
if(!list.contains(a[i])){
list.add(a[i]);
}
}
return (String[])list.toArray(new String[list.size()]);
}