1.问题描述
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.
来自 https://leetcode.com/problems/reverse-linked-list-ii/description/
2.题目分析
题目分析:链表中区间为[m,n]的元素进行反转,因此我们先定位到这个区间(leftpre,right],使用206题的方法进行反转。最后进行两个链表的拼接。
3.c++代码
//我的代码:(beats 100%)
ListNode* reverseBetween(ListNode* head, int m, int n)
{
if (head == NULL)return head;
ListNode*dummy = new ListNode(0);
dummy->next = head;
ListNode*leftpre = dummy;
ListNode*right = dummy;
for (int i = m; i <= n; i++)
right = right->next;
for (int j = 1; j < m; j++)
{
leftpre = leftpre->next;
right = right->next;
}
//leftpre->next = right->next;
ListNode*newhead = new ListNode(leftpre->next->val);
newhead->next = right->next;
right->next = NULL;
ListNode*p = leftpre->next->next;
ListNode*tmp;
while (p)
{
tmp = p->next;
p->next = newhead;//头插法
newhead = p;
p = tmp;
}
leftpre->next = newhead;
return dummy->next;
}