首页 > 代码库 > Reverse Linked List II
Reverse Linked List II
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* reverseBetween(ListNode* head, int m, int n) { ListNode* newHead = new ListNode(-1); newHead->next = head; ListNode* p = newHead; n = n - m + 1; while(--m){ p = p->next; } ListNode* cur = p->next; ListNode* ptail = cur; while (n--) { ListNode* pNext = p->next; ListNode* curNext = cur->next; p->next = cur; cur->next = pNext; cur = curNext; ptail->next = curNext; } return newHead->next; }};
Reverse Linked List II
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。