LeetCode - Swap Nodes in Pairs

Swap Nodes in Pairs

2013.12.7 01:54

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.

Solution:

　　This kind of problem is designed just to make some trouble for you, and the interviewer must be very glad watching you getting into a bloody mess with the code. So let's first write down the process in language description:

　　　　1. use two pointers $p1/$p2 pointing to the first/second node in each pair.

　　　　2. change the $next pointers of p1 and p2

　　　　3. swap $p1 and $p2, make sure $p1 is still before $p2.

　　　　4. record current $p2 as the parent pointer $par, which will be used in the next pair.

　　　　5. move both $p1 and $p2 forward by two steps, don't forget to check for nullptr.

　　　　6. {do the swapping for the next pair}, and point $par->next to $p1

　　Time complexiy is O(n), space complexity is O(1). The real problem is whether you can write the code nice and clean. Try to be careful and patient.

Accepted code:

// This problem is tricky, 3WA
/**
 * 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.
        ListNode *p1 = nullptr, *p2 = nullptr;
        
        if(head == nullptr){
            return head;
        }
        
        p1 = head;
        p2 = head->next;
        if(p2 == nullptr){
            return p1;
        }
        
        bool first_swap = true;
        ListNode *tmp, *par;
        ListNode *root;
        
        par = new ListNode(0);
        root = par;
        par->next = p1;
        while(true){
            p1->next = p2->next;
            p2->next = p1;
            
            tmp = p1;
            p1 = p2;
            p2 = tmp;
            
            par->next = p1;
            par = p2;
            
            if(first_swap){
                head = p1;
                first_swap = false;
            }
            
            p1 = p1->next;
            p2 = p1->next;
            if(p2 == nullptr){
                break;
            }
            
            p1 = p1->next;
            p2 = p1->next;
            if(p2 == nullptr){
                break;
            }
        }
        
        delete root;
        root = nullptr;
        return head;
    }
};