【一天一道LeetCode】#24. Swap Nodes in Pairs

一天一道LeetCode系列

（一）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.

（二）解题

给定一个链表，交换相邻两个节点，这道题要特别注意越界问题。

评级easy的题！就不多废话了。

/** * 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) return NULL;        ListNode* p = head;        ListNode* pnext = head->next;        while(p!=NULL&&pnext!=NULL)        {            int tmp = pnext->val;            pnext->val = p->val;            p->val = tmp;            if(pnext->next !=NULL) p = pnext->next; //考虑越界问题,如[1,2,3,4]            else p=NULL;            if(p!= NULL && p->next != NULL) pnext = p->next;//考虑越界问题，如[1,2,3,4,5]            else pnext=NULL;        }        return head;    }};