AI代码解释 publicclassSolution{publicListNodereverseBetween(ListNode head,int m,int n){if(m==n||head==null||head.next==null){returnhead;}ListNode pre=newListNode(0);pre.next=head;ListNode Mpre=pre;ListNode NodeM=head;ListNode NodeN=head;int mNum=1;int nNum=1;while(mNum<m&&NodeM!=nu...
1publicclassSolution {2publicListNode reverseBetween(ListNode head,intm,intn) {3if(m >= n || head ==null) {4returnhead;5}67ListNode dummy =newListNode(0);8dummy.next =head;9head =dummy;1011for(inti = 1; i < m; i++) {12if(head ==null) {13returnnull;14}15head =head.next;...
Lintcode35 Reverse Linked List solution 题解 【题目描述】 Reverse a linked list. 翻转一个链表 【题目链接】 http://www.lintcode.com/en/problem/reverse-linked-list/ 【题目解析】 这题要求我们翻转[m, n]区间之间的链表。对于链表翻转来说,几乎都是通用的做法,譬如p1 -> p2 -> p3 -> p4,如果我...
Reverse a linked list from position m to n. Notice:Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list. 翻转链表中第m个节点到第n个节点的部分 注意:m,n满足1 ≤ m ≤ n ≤ 链表长度 【题目链接】 http://www.lintcode.com/en/problem/reverse-linked-list-ii/ 【...
代码 迭代的方式处理 // 206. Reverse Linked List // https://leetcode.com/problems/reverse-linked-list/description/ // 时间复杂度: O(n) // 空间复杂度: O(1) class Solution { public: ListNode* reverseList(ListNode* head) { ListNode* pre = NULL; ListNode* cur = head; while(cur != ...
* }*/classSolution {public:/** * @param head: The first node of linked list. * @return: The new head of reversed linked list.*/ListNode*reverse(ListNode *head) {//case1: empty listif(head == NULL)returnhead;//case2: only one element listif(head->next == NULL)returnhead;//cas...
* Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ classSolution{ publicListNodereverseList(ListNode head){ if(head ==null|| head.next ==null) { ...
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode reverseKGroup(ListNode head, int k) { if (head == null || head.next == null) return head; int count =...
* Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */classSolution{public:ListNode*reverseList(ListNode*head){ListNode*next=NULL;ListNode*pre=NULL;if(head==NULL){returnNULL;}while(head!=NULL){next...
* Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */classSolution{public:ListNode*reverseList(ListNode*head){//非递归解法// ListNode *prev=head;// ListNode *result=NULL;// ListNode *temp=prev;/...