24. Swap Nodes in Pairs
Explanation:
To solve this problem, we can use a dummy node to help with handling edge cases. We will iterate through the linked list, swapping pairs of adjacent nodes.
- Create a dummy node and point it to the head of the linked list.
- Iterate through the linked list by pairs, swapping the nodes.
- Update the pointers accordingly.
- Return the new head of the linked list after swapping.
Time Complexity: O(N) where N is the number of nodes in the linked list. Space Complexity: O(1)
class Solution {
public ListNode swapPairs(ListNode head) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode current = dummy;
while (current.next != null && current.next.next != null) {
ListNode first = current.next;
ListNode second = current.next.next;
first.next = second.next;
current.next = second;
current.next.next = first;
current = current.next.next;
}
return dummy.next;
}
}Code Editor (Testing phase)
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