24 Swap Nodes in Pairs

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Intuition & Algorithm

Question 25 is a more general question of this one.

Suppose there are three pairs $p_0$, $p_1$ and $p_2$. We need to swap nodes in $p_1$, after that, we should reconnect it to it’s adjacent pairs by:

For the first pair, $p_0(1)=null$ so we don’t need the first equation and and for the last one, $p_2(0)=null$;

Code

  1. Recursion
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    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    class Solution {
        public ListNode swapPairs(ListNode head) {
            return dp(head);
        }
        
        private ListNode dp(ListNode head){
            if(head == null || head.next == null)
                return head;
            ListNode next = head.next.next;
            head.next.next = head;
            head = head.next;
            head.next.next = dp(next);
            return head;
        }
    }

2.Iteration

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null || head.next == null)
            return head;
        ListNode gH = head;
        ListNode gT = head.next;
        head = gT;
        while(gH != null && gT != null){
            ListNode tmp = gT.next;
            gT.next = gH;
            if(tmp != null && tmp.next != null){
                gH.next = tmp.next;
                gH = tmp; gT = tmp.next;
            }
            else{
                gH.next = tmp;
                gH = tmp; gT = null;
            }
        }
        return head;
    }
}

Complexity

Time Complexity: $O(n)$ becasue we iterate over the entire list

Space Complexity: For the recursion solution, $O(\lceil n/2 \rceil)$ space is necessary to store all the pairs. We cost constant space $O(1)$ for the iteration method.