Swap Nodes in Pairs 解法

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第 12 周题目
难度：Media
LeetCode题号：24

题目

Description：

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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思考

整体的思想是维持三个指针变量
current, next, pre，分别表示当前位置，下一个位置和上一个位置
每次循环，current指向next的下一个，next指向current，同时pre指向next（原本pre指向current）
剩下的就是处理开头和结尾的
因为存在pre，所以开始的两个节点调换不写在循环里
最后当next为null时退出循环

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代码

/** * 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 * current,* Next, * pre, * result;        current = head;        Next = current->next;        result = Next;        current->next = Next->next;        Next->next = current;        pre = current;        if (current->next != NULL) {            current = current->next;        }        Next = current->next;        while (Next != NULL) {            current->next = Next->next;            Next->next = current;            pre->next = Next;            pre = current;            if (current->next != NULL) {                current = current->next;            }            Next = current->next;        }        return result;    }};