解码和编码的高难度有关问题

解码和编码的高难度问题？
用javascript对url参数编码     encodeURI(String   args);

怎么用Java的方法对其解码？？？

String   s=request.getParameter( "args ");
Stirng   result=XXX.function(s);

java.net.URLDecoder?这个好象不行？

------解决方案--------------------
s = unescape(s);

public static String escape(String src) {
int i;
char j;
StringBuffer tmp = new StringBuffer();
tmp.ensureCapacity(src.length() * 6);

for (i = 0; i < src.length(); i++) {

j = src.charAt(i);

if (Character.isDigit(j) || Character.isLowerCase(j)
|| Character.isUpperCase(j))
tmp.append(j);
else if (j < 256) {
tmp.append( "% ");
if (j < 16)
tmp.append( "0 ");
tmp.append(Integer.toString(j, 16));
} else {
tmp.append( "%u ");
tmp.append(Integer.toString(j, 16));
}
}
return tmp.toString();
}

public static String unescape(String src) {
StringBuffer tmp = new StringBuffer();
tmp.ensureCapacity(src.length());
int lastPos = 0, pos = 0;
char ch;
while (lastPos < src.length()) {
pos = src.indexOf( "% ", lastPos);
if (pos == lastPos) {
if (src.charAt(pos + 1) == 'u ') {
ch = (char) Integer.parseInt(src
.substring(pos + 2, pos + 6), 16);
tmp.append(ch);
lastPos = pos + 6;
} else {
ch = (char) Integer.parseInt(src
.substring(pos + 1, pos + 3), 16);
tmp.append(ch);
lastPos = pos + 3;
}
} else {
if (pos == -1) {
tmp.append(src.substring(lastPos));
lastPos = src.length();
} else {
tmp.append(src.substring(lastPos, pos));
lastPos = pos;
}
}
}
return tmp.toString();
}