小知识，大挑战！本文正在参与“程序员必备小知识”创作活动。
描述
You are given the head of a linked list, and an integer k.
Return the head of the linked list after swapping the values of the kth node from the beginning and the kth node from the end (the list is 1-indexed).
Example 1:
Input: head = [1,2,3,4,5], k = 2
Output: [1,4,3,2,5]
Example 2:
Input: head = [7,9,6,6,7,8,3,0,9,5], k = 5
Output: [7,9,6,6,8,7,3,0,9,5]
Example 3:
Input: head = [1], k = 1
Output: [1]
Example 4:
Input: head = [1,2], k = 1
Output: [2,1]
Example 5:
Input: head = [1,2,3], k = 2
Output: [1,2,3]
Note:
The number of nodes in the list is n.
1 <= k <= n <= 10^5
0 <= Node.val <= 100
解析
根据题意，就是给出了一个链表，并且给出了一个数字，要求我们用将链表的正向第 k 个值和逆向的第 k 个值进行交换，最后返回新的链表的头节点 head 。其实思路比较简单：
- 初始化 first 和 last 都指向 head
- 因为一开始就是第一个位置，所以遍历 k-1 次，将 first 指向正向的第 k 个节点
- 然后再将 first 赋予一个新的指针 new_pointer ，让 new_pointer 和 last 同时向后进行节点遍历，直到 new_pointer.next 为空，此时所找到的 last 就是逆向的第 k 个节点，将两者的值进行交换即可
- 返回新生成的链表头节点 head
解答
class ListNode(object):
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution(object):
def swapNodes(self, head, k):
"""
:type head: ListNode
:type k: int
:rtype: ListNode
"""
first = last = head
for _ in range(1,k):
first = first.next
new_pointer = first
while new_pointer.next:
last = last.next
new_pointer = new_pointer.next
first.val, last.val = last.val, first.val
return head
运行结果
Runtime: 1610 ms, faster than 33.72% of Python online submissions for Swapping Nodes in a Linked List.
Memory Usage: 92.8 MB, less than 22.87% of Python online submissions for Swapping Nodes in a Linked List.
解析
从上面可以启发我们其实关键就在于找正向和逆向的第 k 个位置，然后进行值的交换，这可以利用列表进行操作：
- 用 head 赋给 current ，表示一个指向当前节点的变量
- 然后使用 current 不断向下进行遍历，按顺序将遍历到的每个节点的索引存入列表 nodes 中
- 将 nodes 中索引为 k-1 的节点和索引为 len(nodes)-k 的节点的值进行交换
- 最后返回 head 即可
解答
class ListNode(object):
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution(object):
def swapNodes(self, head, k):
"""
:type head: ListNode
:type k: int
:rtype: ListNode
"""
nodes = []
current = head
while current:
nodes.append(current)
current = current.next
nodes[k-1].val, nodes[len(nodes)-k].val = nodes[len(nodes)-k].val, nodes[k-1].val
return head
运行结果
Runtime: 2043 ms, faster than 20.53% of Python online submissions for Swapping Nodes in a Linked List.
Memory Usage: 91.3 MB, less than 49.27% of Python online submissions for Swapping Nodes in a Linked List.
原题链接：leetcode.com/problems/sw…
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