反转链表

题目

Given a singly-linked list, reverse the list. This can be done iteratively or recursively. Can you get both solutions?

Example:
Input: 4 -> 3 -> 2 -> 1 -> 0 -> NULL
Output: 0 -> 1 -> 2 -> 3 -> 4 -> NULL

要求做出递归法和迭代法2种写法。

分析

递归法：
将第二个元素开始的尾部链表递归反转，然后拼到第一个元素上去。接下来需要额外做一次迭代，找到新链表的头指针以便返回。这里暂时没想到更有效的解法，可能不是最优的。

迭代法：
用2个指针，一个 created 表示新创建的结果链表的头指针，另一个 remaining 表示剩余可用的节点的指针。从原来的链表上依次拆除元素，添加到新链表，添加的时候，顺便调整指针以便反转之。这里，链表调整各节点指向关系的代码不算难，但是有点饶人，需要多加练习，以增加熟练度，并做到思路清楚。

时间复杂度 O(n).

代码

class ListNode(object):
  def __init__(self, x):
    self.val = x
    self.next = None
  
  # Function to print the list
  def printList(self):
    node = self
    output = '' 
    while node != None:
      output += str(node.val)
      output += " "
      node = node.next
    print(output)

  # Iterative Solution
  def reverseIteratively(self, head):
    created = None
    remaining = head

    while remaining:
      new_remaining = remaining.next
      remaining.next = created
      created = remaining
      remaining = new_remaining
    
    return created

  # Recursive Solution      
  def reverseRecursively(self, head):
    if head.next:
      tail = self.reverseRecursively(head.next)
      p = tail
      while p.next:
        p = p.next
      p.next = head
      head.next = None
      return tail
    else:
      return head

# Test Program
# Initialize the test list: 
testHead = ListNode(4)
node1 = ListNode(3)
testHead.next = node1
node2 = ListNode(2)
node1.next = node2
node3 = ListNode(1)
node2.next = node3
testTail = ListNode(0)
node3.next = testTail

print("Initial list: ")
testHead.printList()
# 4 3 2 1 0
testHead.reverseIteratively(testHead)
# testHead.reverseRecursively(testHead)
print("List after reversal: ")
testTail.printList()
# 0 1 2 3 4