coding-interview|September 16, 2019|2 min read

Swap Nodes Pairs in Link List - Leet Code Solution

TL;DR

Iterative pointer manipulation — swap each pair by redirecting next pointers, use a prev pointer to stitch pairs together. O(n) time, O(1) space.

Swap Nodes Pairs in Link List - Leet Code Solution

Problem Statement

Given a linked list, swap every two adjacent nodes and return its head. You may not modify the values in the list’s nodes, only nodes itself may be changed.

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

Solution

  • Lets look at the first two nodes first.
  • We need to move head to second. (One time operation)
  • Move next of First to next of Second
  • Move next of Second to First
  • All looks fine. We can go like this in pairs.
  • Now, what about joining the pairs?
  • We need another pointer who will be responsible of joining these pairs.

Code

public static class ListNode {
	int val;
	ListNode next;
	ListNode(int x) { val = x; }
	@Override
	public String toString() {
		StringBuffer sb = new StringBuffer();
		ListNode t = this;
		while (t != null) {
			sb.append(t.val).append(" -> ");
			t = t.next;
		}
		return sb.toString();
	}
}

public ListNode swapPairs(ListNode head) {
	if (head == null || head.next == null) return head;
	
	ListNode first = head;
	ListNode second = head.next;
	ListNode prev = null;
	
	ListNode result = second;
	while (first != null && second != null) {
		first.next = second.next;
		second.next = first;
		
		if (prev != null) {
			prev.next = second;
		}
		
		prev = first;
		
		first = first.next;
		if (first != null) {
			second = first.next;
		}
	}
	
	return result;
}

Leet Code submission result

Runtime: 0 ms, faster than 100.00% of Java online submissions for Swap Nodes in Pairs. Memory Usage: 34.5 MB, less than 100.00% of Java online submissions for Swap Nodes in Pairs.

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