题目
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.
/**
* 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) {
}
};
思路
代码
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if(head == NULL) {
return NULL;
}
return swapTwoPairs(head);
}
ListNode* swapTwoPairs(ListNode* First) {
if(First->next == NULL) {
return First;
}
if(First->next->next != NULL) {
ListNode* Second = First->next->next;
ListNode* Third = First->next;
First->next->next = First;
First->next = swapTwoPairs(Second);
return Third;
} else {
First->next->next = First;
ListNode* Third = First->next;
First->next = NULL;
return Third;
}
}
};
链表节点交换算法
本文介绍了一种链表中相邻节点两两交换位置的算法实现。该算法仅使用常数空间,通过递归方式改变节点指针指向来完成节点位置的交换,而不需要修改链表中的元素值。
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