算法第五周Swap Nodes in Pairs[medium]

Swap Nodes in Pairs[medium]

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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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Analysis

题目是很容易理解的。就是将相邻的结点交换，并且题目中明确要求不能直接交换结点的值，同时，从题目中我们能够看出他是每两个结点进行一次交换。我采用的方法是每三个结点交换后两个，首先创建一个结点h，它的next指针指向list的head，如果h的下一个和下下个均不为空，我们即可利用h来交换后两个结点，之后更新h的值，是他等于上一次处理的最后一个结点。

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Solution

/** * 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* h = new ListNode(1);        h->next = head;        ListNode* c = h;        ListNode* p = NULL;        ListNode* t = NULL;        while (c->next != NULL &&c->next->next != NULL) {            p = c->next;            t = c->next->next;            p->next = t->next;            c->next = t;            t->next = p;            c = p;        }        return h->next;    }};