LeetCode 24 - Swap Nodes in Pairs

一、问题描述

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

.

Note:

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

给一个链表，交换每两个相邻的结点，返回新链表。

二、解题报告

解法一：操作值域

直接交换结点的

val

是最简单的：

class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if(head == NULL)
return NULL;
ListNode* first = head;
ListNode* second = head->next;
while(first!=NULL && second!=NULL) {
int temp = first->val;      // 交换val
first->val = second->val;
second->val = temp;
if(first->next!=NULL)
first = first->next->next;
if(second->next!=NULL)
second = second->next->next;
}
return head;
}
};

解法二：操作结点

题目要求：You may not modify the values in the list, only nodes itself can be changed. 好吧，那就来操作结点吧。

class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if(head == NULL)
return NULL;
ListNode* first = head;
ListNode* second = head->next;
ListNode* p = new ListNode(0);
head = p;
while(first!=NULL && second!=NULL) {
ListNode* temp1 = first;
ListNode* temp2 = second;
if(second->next!=NULL)
second = second->next->next;
if(first->next!=NULL)
first = first->next->next;
p->next = temp2;
p->next->next = temp1;
p = p->next->next;
p->next = NULL;
}
if(first!=NULL) {
p->next = first;
}
return head->next;
}
};

空间复杂度为O(1)。

LeetCode答案源代码：https://github.com/SongLee24/LeetCode