LeetCode - Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

For example,

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

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

http://oj.leetcode.com/problems/first-missing-positive/

Solution:

If length is 0 or 1, return head. assume there is ...->node1->node2->node3->..., what we want is ...->node2->node1->node3->...

https://github.com/starcroce/leetcode/blob/master/swap_nodes_in_pairs.cpp

// 24 ms for 55 test cases
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(head == NULL || head->next == NULL){
            return head;
        }
        ListNode *node1 = head;
        ListNode *node2 = node1->next;
        // prev is the last node of the swapped list
        ListNode *prev = NULL;
        while(node2 && node2->next){
            ListNode *tmp = node2->next;
            node2->next = node1;
            node1->next = tmp;
            // if at the start
            if(prev == NULL){
                head = node2;
                prev = node1;
            }
            else{
                prev->next = node2;
                prev = node1;
            }
            node1 = node1->next;
            node2 = node1->next;
        }
        // length = 2
        if(prev == NULL){
            head = node2;
            node2->next = node1;
            node1->next = NULL;
            return head;
        }
        // length is odd
        if(node2 == NULL){
            return head;
        }
        // the last swap
        node2->next = node1;
        node1->next = NULL;
        prev->next = node2;
        return head;
    }
};