[LeetCode] Swap Nodes in Pairs

题目：

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.

解答：

/**
 * 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) {
        //如果没有节点或只有一个节点
        if(head == NULL || head -> next == NULL) {
            return head;
        }
        else {
            ListNode *first, *second, *newHead = NULL, *pre = NULL;
            first = head;
            //second = head -> next;
            while(first != NULL && first -> next != NULL) {
                second = first -> next;
                
                if(newHead == NULL) {
                    newHead = second;
                }
                
                first -> next = second -> next;
                second -> next = first;
                if(pre != NULL) {
                    pre -> next = second;
                }
                
                pre = first;
                first = first -> next;
            }
            return newHead;
        }
    }
};

思路：

这道题没什么难度，容易忽略的就是只使用两个指针，即指向需要调换的两个元素，这样做是不够的，比如1→2→3→4，第一次调换为2→1→3→4，这时候两个指针应该指向3和4了，但是3、4调换完毕后，1的next仍指向3，导致错误，所以需要第三个指针，指向当前需要调换的两个元素的前一个元素，以便做好next的设置。