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;    }};