- Reverse Linked List II My Submissions QuestionEditorial Solution
Total Accepted: 70579 Total Submissions: 252986 Difficulty: Medium
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.
思路:
时间复杂度:O(n),空间复杂度O(1)
找到指定段元素,对该段元素逆转,同时保留指定段的两端的邻居
/**
* 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) {
assert(n>=m);
if(m==n)return head;
int i=1,j=1;
ListNode *p=head,*q=head,*prep=NULL;//保留指定段首元素的前一元素
while(p!=NULL&&i++<m){
if(i==m)prep=p;
p=p->next;
}
while(q!=NULL&&j++<n)q=q->next;/找到各自的最终位置,越界为空
if(p==NULL)return head; //起始位置越界,直接返回head
ListNode * tmp = p->next,*prenode = p,*qnext=q->next;
while(tmp!=qnext) //还未到达指定段的下一元素,则当前元素指向前一元素
{
ListNode *nextnode = tmp->next;
tmp->next = prenode;
prenode = tmp;
tmp = nextnode;
}
p->next = qnext;
if(prep!=NULL)prep->next = prenode;
else head = prenode;
return head;
}
};