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Formatted question description: https://leetcode.ca/all/1721.html

1721. Swapping Nodes in a Linked List

Level

Medium

Description

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 k-th node from the beginning and the k-th node from the end (the list is 1-indexed).

Example 1:

Image text

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 <= 10^5
  • 0 <= Node.val <= 100

Solution

Use pointers to obtain the two nodes, which are node1 and node2. The node node1 is the k-th node starting from the beginning, and the node node2 is the k-th node from the end. Then obtain the two nodes’ values and swap their values.

  • /**
 * 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) {
        ListNode node1 = head;
        for (int i = 1; i < k; i++)
            node1 = node1.next;
        ListNode node2 = head, temp = node1;
        while (temp.next != null) {
            node2 = node2.next;
            temp = temp.next;
        }
        int val1 = node1.val, val2 = node2.val;
        node1.val = val2;
        node2.val = val1;
        return head;
    }
}

############

/**
 * 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) {
        ListNode fast = head;
        while (--k > 0) {
            fast = fast.next;
        }
        ListNode p = fast;
        ListNode slow = head;
        while (fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }
        ListNode q = slow;
        int t = p.val;
        p.val = q.val;
        q.val = t;
        return head;
    }
}

  • // OJ: https://leetcode.com/problems/swapping-nodes-in-a-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
    int getLength(ListNode *head) {
        int ans = 0;
        for (; head; head = head->next) ++ans;
        return ans;
    }
    ListNode *getNode(ListNode *head, int k) {
        while (--k) head = head->next;
        return head;
    }
public:
    ListNode* swapNodes(ListNode* head, int k) {
        int len = getLength(head), r = len - k + 1;
        if (r == k) return head;
        ListNode *a = getNode(head, k), *b = getNode(head, r);
        swap(a->val, b->val);
        return head;
    }
};

  • # 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:
        fast = head
        for _ in range(k - 1):
            fast = fast.next
        p = fast
        slow = head
        while fast.next:
            slow, fast = slow.next, fast.next
        q = slow
        p.val, q.val = q.val, p.val
        return head

############

# 1721. Swapping Nodes in a Linked List
# https://leetcode.com/problems/swapping-nodes-in-a-linked-list/

# 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):
        res = []
        
        curr = head
        while curr:
            res.append(curr)
            curr = curr.next
        
        res[k-1].val, res[-k].val = res[-k].val, res[k-1].val
        
        return head

  • /**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func swapNodes(head *ListNode, k int) *ListNode {
	fast := head
	for k > 1 {
		fast = fast.Next
		k--
	}
	p := fast
	slow := head
	for fast.Next != nil {
		slow, fast = slow.Next, fast.Next
	}
	q := slow
	p.Val, q.Val = q.Val, p.Val
	return head
}

  • /**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function swapNodes(head: ListNode | null, k: number): ListNode | null {
    let fast = head;
    while (--k) {
        fast = fast.next;
    }
    let p = fast;
    let slow = head;
    while (fast.next) {
        slow = slow.next;
        fast = fast.next;
    }
    let q = slow;
    [p.val, q.val] = [q.val, p.val];
    return head;
}


  • /**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode SwapNodes(ListNode head, int k) {
        ListNode fast = head;
        while (--k > 0) {
            fast = fast.next;
        }
        ListNode p = fast;
        ListNode slow = head;
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        ListNode q = slow;
        int t = p.val;
        p.val = q.val;
        q.val = t;
        return head;
    }
}

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