Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given1->2->3->4->5->NULL, m = 2 and n = 4,
return1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
首先找出翻转开始的点,主要喜欢出错的地方就是找到开始的点之后如何一步一步翻转。
每次循环翻转一次,有点不太好表达,忘记的时候来看看代码。
/** * 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) { if(!head || m<1 || n<1 || m>n) return nullptr; ListNode temp(-1); temp.next=head; ListNode *dummy=&temp,*cur=dummy,*start=dummy; for(int i=0;i<m;i++) { if(!cur) return nullptr; start=cur; cur=cur->next; } for(int i=0;i<n-m;i++) { ListNode *t=cur->next; cur->next=t->next; t->next=start->next; start->next=t; } return dummy->next; } };