[LeetCode92]Reverse Linked List II

题目：

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

.

反转m到n处的单链表

The basic idea is as follows:

(1) Create a

new_head

that points to

head

and use it to locate the immediate node before the

m

-th (notice that it is

1

-indexed) node

pre

;

(2) Set

cur

to be the immediate node after

pre

and at each time move the immediate node after

cur

(named

move

) to be the immediate node after

pre

. Repeat it for

n - m

times.

分类：Linked List

代码：

class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
ListNode* new_head = new ListNode(0);
new_head -> next = head;
ListNode* pre = new_head;
for (int i = 0; i < m - 1; i++)
pre = pre -> next;
ListNode* cur = pre -> next;
for (int i = 0; i < n - m; i++) {
ListNode* move = cur -> next;
cur -> next = move -> next;
move -> next = pre -> next;
pre -> next = move;
}
return new_head -> next;
}
};