【Leetcode】Reversed Linked List

【题目】

Reverse a singly linked list.

Hint:

A linked list can be reversed either iteratively or recursively. Could you implement both?

【思路】

我自己想的办法就是可以用一个stack，先把每一个元素都装进去，然后在倒着吐出来。

放在新的list里面。

需要两个指针，一个指向新的list的root，另一个走一遍新list.

【代码】自己提交的代码，一遍过

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseList(ListNode head) {
if(head == null) return null;
Stack<Integer> stack = new Stack<Integer>();
while(head!=null){
stack.push(head.val);
head = head.next;
}
System.out.println(stack);
ListNode newRoot = new ListNode(0);
ListNode p = new ListNode(stack.pop());

newRoot.next = p;
while(!stack.isEmpty()){
p.next = new ListNode(stack.pop());
p = p.next;

}
return newRoot.next;

}
}

[Other’s code]

Iterative ! !

public class Solution {
public ListNode reverseList(ListNode head) {
if(head == null) return head;

ListNode next = head.next;
head.next = null;

while(next != null){
ListNode temp = next.next;
next.next = head;
head = next;
next = temp;
}
return head;
}
}

Recursive!!!!

public class Solution {
public ListNode reverseList(ListNode head) {
if(head == null) return head;
ListNode next = head.next;
head.next = null;

return recursive(head,next);
}

private ListNode recursive(ListNode head, ListNode next){
if(next == null)    return head;
ListNode temp = next.next;
next.next = head;
return recursive(next,temp);

}
}