92. 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.

反转特定区域的链表

/**
 * 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) {
        ListNode *dummy = new ListNode(-1);
        dummy->next = head;

        ListNode *prev = dummy;
        //将链表移动到第m-1个节点
        for(int i = 1; i < m; i++)
            prev = prev->next;

        ListNode *cur = prev->next;
        for(int i = 0; i  < n- m; i++){
            ListNode *after = cur->next;
            cur->next = after->next;
            after->next = prev->next;
            prev->next = after;
        }
        return dummy->next;
    }
};