反转单链表:迭代、递归

【问题】给定单链表的头节点 head ，请反转链表，并返回反转后的链表的头节点。

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

法一：迭代

（1）Python

#Python
class Solution(object):
    def reverseList(self, head):
        pre = None
        while head is not None:
            next = head.next
            head.next = pre 
            pre = head
            head = next
        return pre



class Solution(object):
    def reverseList(self, head):
        prev, curr = None,head
        while curr is not None:
            next = curr.next
            curr.next = prev
            prev = curr
            curr = next
        return prev

（2）Java

//Java
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null){
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
}

方法二：递归

（1）大问题拆成两个子问题）

（2）子问题的求解方式和大问题一样

（3）存在最小子问题

(1)Python 

#Python
class Solution(object):
    def reverseList(self, head):
        #递归终止条件
        if head is None or head.next is None:
            return head
        p = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return p

（2）Java

//java
class Solution {
    public ListNode reverseList(ListNode head) {
        // 1. 递归终止条件
        if(head == null || head.next == null){
            return head;
        }
        ListNode p = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return p;
       
    }
}