leetcode-algorithms-24 Swap Nodes in Pairs
leetcode-algorithms-24 Swap Nodes in Pairs
Given a linked list, swap every two adjacent nodes and return its head.
Example:
Given 1->2->3->4, you should return the list as 2->1->4->3.
Note:
Your algorithm should use only constant extra space.
You may not modify the values in the list's nodes, only nodes itself may be changed.
解法
一个链表指针指向链表的首地址,这个用来改变node的值,且将指针后移.
再申明两个链表指针,其中一个指针b是另个指针a的下个值(b = a->next).
现在循环处理.
/**
* 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 **l = &head;
ListNode *a = nullptr;
ListNode *b = nullptr;
while((a = *l) && (b = a->next))
{
a->next = b->next;
b->next = a;
*l = b;
l = &(a->next);
}
return head;
}
};
时间复杂度: O(n).
空间复杂度: O(1).
