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This approach uses iteration to systematically swap each pair of adjacent nodes in the linked list. The main idea is to traverse the list while keeping track of pointers to the nodes before and after the pair being swapped. By adjusting these pointers, we effectively swap nodes without modifying their values.
Time Complexity: O(n), where n is the number of nodes in the linked list. We process each node once.
Space Complexity: O(1), We are using constant extra space.
1var swapPairs = function(head) {
2 let dummy = new ListNode(0);
3 dummy.next = head;
4 let prev = dummy;
5
6 while (prev.next !== null && prev.next.next !== null) {
7 let a = prev.next;
8 let b = a.next;
9 a.next = b.next;
10 b.next = a;
11 prev.next = b;
12 prev = a;
13 }
14
15 return dummy.next;
16};The JavaScript implementation follows the same pattern by using a dummy node for simplified pointer/structure manipulation to swap pairs of nodes. We readjust pointers within the loop, handling swap in pairs of nodes and ensuring each step maintains proper linkage.
This code ensures that, even when the number of nodes is odd, the remaining node (if any) is left in place without a pair to swap with.
This approach leverages recursion to swap every two adjacent nodes. The key idea is to break the problem into subproblems where you solve for a pair and recursively call to solve for the next. The recursion naturally handles the base case where less than two nodes remain to be swapped.
Time Complexity: O(n), every node is processed once.
Space Complexity: O(n), the call stack in recursive functions uses space proportional to the number of elements.
1
The Python function checks if there are fewer than two nodes left, returning the head. It takes head.next, which is the new head in the swapped pair, and performs a recursive call on the list starting after the next node, setting the returned result to head.next.
This recursive setup effectively 'switches' the pairs and uses the call stack to finish the swaps for the entire list.