原題

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:

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

Reference Answer

思路分析

解題方法有很多種：

1. 藉助python list，儲存每個節點，再反向遍歷，重新連結即可。（時間複雜度 O(n)，空間複雜度 O(n)）
2. 進一步優化，直接在進行遍歷的時候，儲存當前節點，進行反向連結即可，具體看參考程式碼（此時，時間複雜度O(n)，空間複雜度O(1)）

Code One
時間複雜度 O(n)，空間複雜度 O(n)

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseList(self,
 
 head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        temp_list = []
        while head:
            end = head
            temp_list.append(head)
            head = head.next
        root = end
        for
 
 index in range(len(temp_list)-2, -1, -1):
            end.next = temp_list[index]
            end = end.next
        end.next = None
        return root

Code Two
時間複雜度 O(n)，空間複雜度 O(1)

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        
        real_end = end = head
        # end.next = None
        head = head.next
        while head:
            temp = head
            head = head.next
            temp.next = end
            end = temp
        real_end.next = None
        return end    
        

Note:

• 注意方法二中，最終要設定 real_end.next = None