[LeetCode]24. Swap Nodes in Pairs

24. 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.

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// 事先分析好需要几个临时指针变量// 画图分析一下指针变换顺序即可/** * 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) {        ListNode* dummyHead = new ListNode(0);        // 指向dummyHead的指针        ListNode* d_pCur = dummyHead;        dummyHead->next = head;        ListNode* pCur = head;        // 至少两个非空结点才swap        while (pCur && pCur->next) {            ListNode* temp = pCur->next;            ListNode* pNext = pCur->next;            pCur->next = pCur->next->next;            pNext->next = pCur;            d_pCur->next = temp;            d_pCur = d_pCur->next->next;            pCur = pCur->next;        }        return dummyHead->next;    }};