Swap Nodes in Pairs 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...
*/classSolution{publicListNodeswapPairs(ListNode head){ListNodedummy=newListNode(-1); dummy.next = head;ListNodecurr=dummy;while(curr.next !=null&& curr.next.next !=null) {ListNodefirst=curr.next;ListNodesecond=curr.next.next;// swap two nodesfirst.next = second.next; second.next = first; ...
给你一个链表,两两交换其中相邻的节点,并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题(即,只能进行节点交换)。 示例1: 输入:head = [1,2,3,4]输出:[2,1,4,3] 示例2: 输入:head = []输出:[] 示例3: 输入:head = [1]输出:[1] ...
24. Swap Nodes in Pairs Medium Topics Companies Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.) Example 1: Input: head = [1,2,3,4] ...
leetcode 24. Swap Nodes in Pairs 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 ...
Leetcode 24.Swap Nodes in Pairs 给你一个链表,交换相邻两个节点,例如给你 1->2->3->4,输出2->1->4->3。 我代码里在head之前新增了一个节点newhead,其实是为了少写一些判断head的代码。 public class Solution { public ListNode swapPairs(ListNode head) {...
*/class Solution{public:ListNode*swapPairs(ListNode*head){if(!head||!head->next){returnhead;}ListNode*latter=head->next;head->next=swapPairs(latter->next);latter->next=head;returnlatter;}}; Reference https://leetcode.com/problems/swap-nodes-in-pairs/description/...
总结: Linkedlist, swap nodes ** Anyway, Good luck, Richardo! 其实就是 Reverse Nodes in k-Group - http://www.jianshu.com/p/25f5269c729d 的特殊情况。 思路很简单,没什么好说的。 My code: /** * Definition for singly-linked list. ...
* int val; * ListNode next; * ListNode(int x) { val = x; } * } */publicclassSolution{/** * @param head a ListNode * @return a ListNode */publicListNodeswapPairs(ListNodehead){ListNodedummy=newListNode(0);dummy.next=head;head=dummy;while(head.next!=null&&head.next.next!=null){Li...
Swap Nodes in Pairs - LeetCode注意点考虑链表为空 解法解法一:维护三个指针,前中后,调换这三个位置的next指针即可。时间复杂度O(n)/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class ...