LeetCode 24 Swap Nodes in Pairs (C,C++,Java,Python)

Problem:

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.

Solution:

单纯的链表指针操作，没有什么好讲的，复杂度O(n)

题目大意：

给一个链表，要求将链表的奇数个节点和它后边的节点进行互换，并且不能更改节点的值（也就是不能互换节点的值），空间复杂度要求为O(1)

Java源代码（272ms）：

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode swapPairs(ListNode head) {
ListNode p=new ListNode(0),s;
p.next=head;
head=p;
while(p.next!=null && p.next.next!=null){
s=p.next.next;
p.next.next=s.next;
s.next=p.next;
p.next=s;
p=s.next;
}
return head.next;
}
}

C语言源代码（1ms）不开辟新的节点内存：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
struct ListNode* swapPairs(struct ListNode* head) {
struct ListNode *p=head,*s;
if(p!=NULL && p->next!=NULL){
s=p->next;
p->next=s->next;
s->next=p;
head=s;
while(p->next!=NULL && p->next->next!=NULL){
s=p->next->next;
p->next->next=s->next;
s->next=p->next;
p->next=s;
p=s->next;
}
}
return head;
}

C++源代码（9ms）：

/**
* 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 *p,*s;
p=new ListNode(0);
p->next=head;
head=p;
while(p->next!=NULL && p->next->next!=NULL){
s=p->next->next;
p->next->next=s->next;
s->next=p->next;
p->next=s;
p=s->next;
}
return head->next;
}
};

Python源代码（81ms）：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
# @param {ListNode} head
# @return {ListNode}
def swapPairs(self, head):
p=ListNode(0)
p.next=head;head=p
while p.next!=None and p.next.next!=None:
s=p.next.next
p.next.next=s.next
s.next=p.next
p.next=s
p=s.next
return head.next