Question

LinkedList 中用递归的方法求最大/小数值.急,谢谢!!!!!!!!!!!!!!!!!!!!!!!

public
class MinMax
{

public static void main(String[]args)
{

MinMax mm = new MinMax();
LinkedList<Integer> list= new LinkedList<Integer>();

public int getMin(LinkedList<Integer> list)
{
求最小值.此处求解..........

}

public int getMax(LinkedList<Integer> list)
{
我自己类推.

}
}

谢谢!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

public static void main(String[] args){
LinkedList list = new LinkedList();
list.add(5);
list.add(4);
list.add(9);
list.add(0);
list.add(1000);
list.add(7000);
System.out.println(LinkedListMax(list, 0));
}
public static int LinkedListMax(LinkedList<?> list, int max) {
if (list.isEmpty())
return max;
int tmp = Integer.parseInt(list.pop().toString());
return max > tmp ? LinkedListMax(list, max) : LinkedListMax(list, tmp);
}
你换成求最小,不用我说吧
但我只想说你这根本没必要啊,LinkedList本身就有个迭代为什么还要来用递归呢

Answer 2

public int getMin(LinkedList<Integer> list)
{
Integer head = list.removefirst();
Integer retval = head;
Integer minRest = getMin(list);

if (retval > minRest)
retval = minRest;
list.offerfirst(head);
return retval;

}不过这题目用递归真心不划算⋯⋯

追问

谢谢您好,请问哪行用了递归?

追答

这两行——
Integer head = list.removefirst();
。。。

Integer minRest = getMin(list);

Answer 3

链表怎么可能递归