leetcode解题: Swap Nodes in Pairs (24)

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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.

解法1： O(N), 基本的LinkedList操作

调整后头节点不确定，用dummy node解决。然后就是很普通的node之间的对换。 1 -> 2 -> 3 -> 4, 在换两个node之前，要记录剩下的list的头节点。 换好之后（假设两个node分别为first和second，那么就把first指向剩下的list的头节点)
C++

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/**
 * 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) {
        if (head == NULL || head->next == NULL) {
            return head;
        }
        
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        
        ListNode* ptr = dummy;
        while (ptr->next != NULL && ptr->next->next != NULL) {
            ListNode* first = ptr->next;
            ListNode* second = first->next;
            ListNode* remaining = second->next;
            ptr->next = second;
            second->next = first;
            first->next = remaining;
            ptr = first;
        }
        
        ListNode* res = dummy->next;
        delete dummy;
        return res;
    }
};

Java

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