[Leetcode] Swap Nodes in Pairs

Swap Nodes in Pairs 题解

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题目来源：https://leetcode.com/problems/swap-nodes-in-pairs/description/

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Description

Given a linked list, swap every two adjacent nodes and return its head.

Example

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.

Solution

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if (head == NULL)
            return NULL;
        ListNode *tempHead = new ListNode(0);
        tempHead -> next = head;
        ListNode* front = tempHead;
        ListNode *node1, *node2;
        while (front -> next && front -> next -> next) {
            node1 = front -> next;
            node2 = node1 -> next;
            front -> next = node2;
            node1 -> next = node2 -> next;
            node2 -> next = node1;
            front = node1;
        }
        ListNode* res = tempHead -> next;
        delete tempHead;
        return res;
    }
};

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解题描述

这道题题意是将一个链表中每相邻的一对节点交换位置（已经交换过位置的不再交换）。主要的想法是，每次有三个指针front，node1，node2，在链表中是以front -> node1 -> node2的顺序排列，这里主要要做的就是交换node1和node2的位置，使这段链表变成front -> node2 -> node1，之后将node1赋值给front即可开始下一步交换。

另外这里用到了一个临时链表头tempHead来指向原来的head，便于给front赋初始值，并且在返回新链表头的时候可以快速找到。