[LeetCode] 1721. Swapping Nodes in a Linked List

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]

Constraints:

• The number of nodes in the list is n.
• 1 <= k <= n <= 105
• 0 <= Node.val <= 100

交换链表中的节点。

给你链表的头节点 head 和一个整数 k 。交换链表正数第 k 个节点和倒数第 k 个节点的值后，返回链表的头节点（链表 从 1 开始索引）。

由于只需要swap两个节点的node.val，所以这道题简单很多。这里我提供一个双指针的思路。首先我们计算一下链表的长度len，得到这个长度之后，因为要交换的节点是正数第 K 个和倒数第 K 个节点，其中倒数第 K 个节点正数的index = len + 1 - k。依据这个信息，我们用两个指针从head节点开始遍历，分别在两个被swap的节点停下，swap他们的节点值即可。

时间O(n)

空间O(1)

Java实现

/**
 * 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) {
        int len = 0;
        ListNode cur = head;
        while (cur != null) {
            cur = cur.next;
            len++;
        }

        int first = k;
        int second = len + 1 - k;
        int i = 1;
        ListNode p = head;
        while (p != null && i != first) {
            p = p.next;
            i++;
        }

        int j = 1;
        ListNode q = head;
        while (q != null && j != second) {
            q = q.next;
            j++;
        }

        int temp = p.val;
        p.val = q.val;
        q.val = temp;
        return head;
    }
}

 

LeetCode 题目总结