LeetCode-in-Dart
24. Swap Nodes in Pairs
Medium
Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list’s nodes (i.e., only nodes themselves may be changed.)
Example 1:
Input: head = [1,2,3,4]
Output: [2,1,4,3]
Example 2:
Input: head = []
Output: []
Example 3:
Input: head = [1]
Output: [1]
Constraints:
- The number of nodes in the list is in the range
[0, 100]. 0 <= Node.val <= 100
Solution
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode? next;
* ListNode([this.val = 0, this.next]);
* }
*/
class Solution {
ListNode? swapPairs(ListNode? head) {
if (head == null) {
return null;
}
int len = getLength(head);
return reverse(head, len);
}
// Function to get the length of the linked list
int getLength(ListNode? curr) {
int cnt = 0;
while (curr != null) {
cnt++;
curr = curr.next;
}
return cnt;
}
// Recursive function to reverse pairs of nodes
ListNode? reverse(ListNode? head, int len) {
// Base case: if less than 2 nodes, no need to swap
if (len < 2) {
return head;
}
ListNode? curr = head;
ListNode? prev;
ListNode? next;
prev = null;
// Reverse the first two nodes
for (int i = 0; i < 2; i++) {
next = curr!.next;
curr.next = prev;
prev = curr;
curr = next;
}
// Recursively reverse the rest of the list
head!.next = reverse(curr, len - 2);
return prev;
}
}