python3 解法, 执行用时: 696 ms, 内存消耗: 48.2 MB, 提交时间: 2022-11-28 13:45:38

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def swapNodes(self, head: ListNode, k: int) -> ListNode:
        p,q,n=head,head,head
        i=1
        while n:
            if i<k:
                p=p.next#正数第k个
            if i>k:
                q=q.next#倒数第k个
            n=n.next
            i+=1
        p.val,q.val=q.val,p.val
        return head

java 解法, 执行用时: 2 ms, 内存消耗: 56.1 MB, 提交时间: 2022-11-28 13:43:50

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode swapNodes(ListNode head, int k) {
        // 模拟指针，用来遍历链表
        ListNode cur = head;
        // 用来定位正数第k个节点
        ListNode first = head;
        // 用来定位倒数第k个节点
        ListNode last = head;
        // 用于节点的计数，和节点值的交换
        int count = 1;
        while (cur.next != null) {
            // 找到正数第k个节点
            if (count < k) {
                first = first.next;
            // 找到倒数第k个节点
            } else {
                last = last.next;
            }
            count++;
            cur = cur.next;
        }
        // 交换正数第k个节点和倒数第k个节点的值
        count = first.val;
        first.val = last.val;
        last.val = count;
        return head;
    }
}

python3 解法, 执行用时: 672 ms, 内存消耗: 48.3 MB, 提交时间: 2022-11-28 13:43:32

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def swapNodes(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        slow = fast = head
        for _ in range(k-1):
            fast = fast.next
        pre = fast
        while fast.next:
            slow = slow.next
            fast = fast.next
        pre.val, slow.val = slow.val, pre.val
        return head

python3 解法, 执行用时: 728 ms, 内存消耗: 48.3 MB, 提交时间: 2022-11-28 13:42:59

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def swapNodes(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        ref = list()
        while head:
            ref.append(head)
            head = head.next
        ref[k-1].val, ref[-k].val = ref[-k].val, ref[k-1].val
        return ref[0]