# 1721. Swapping Nodes in a Linked List
###### tags: `Leetcode` `Medium` `Linked List`
Link: https://leetcode.com/problems/swapping-nodes-in-a-linked-list/description/
## 思路
One pass解法
首先找到从```head```开始的第k个元素
然后我们会发现从找到的这个元素到```head```的距离 刚好就是linkedlist的尾巴到'从尾巴往前数的第k个元素'
因此我们新建一个```temp```等于```head``` 然后让```temp```和```curr```一起动 直到```curr```已经到了linked list的尾巴
这时候只有交换两个node的值即可
## Code

class Solution:
    def swapNodes(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        curr = head
        for i in range(k-1):
            curr = curr.next
        fromBeginCurr = curr
        
        temp = head
        while curr.next!=None:
            temp = temp.next
            curr = curr.next
        fromEndCurr = temp

        fromBeginCurr.val, fromEndCurr.val = fromEndCurr.val, fromBeginCurr.val
        
        return head