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[leetcode] 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.
思路:
一般链表反向的题目都是从前到后逐一反向,这个也不例外。bfore指向m的前一个节点(如果m=1,before=NULL)。start就是第m个节点,也就是开始反向的节点。end与after节点逐渐向后变化,end是after前面的那个节点,将after的next指向end。这两个节点的关系靠p来过渡。
题解:
/** * 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(m==n) return head; ListNode *p = head; ListNode *before, *start, *end, *after; before=start=end=after=NULL; for(int i=1;i<m;i++) { before = p; p = p->next; } start=end=p; p = p->next; for(int i=m;i<n;i++) { after = p->next; p->next = end; end = p; p = after; } start->next = after; if(before!=NULL) before->next = end; else head = end; return head; }};
[leetcode] Reverse Linked List II
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