[LeetCode92]Reverse Linked List II

タイトル:
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 .
mからnまでの単一チェーンテーブルを反転
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.
分類:Linked List
コード:

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;
    }
};