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.
分析：头插法。时间复杂度O(n)，空间复杂度O(1)。
class Solution { public: ListNode *reverseBetween(ListNode *head, int m, int n) { if(head == NULL || head->next == NULL || m == n) return head; ListNode *dummy = new ListNode(-1); dummy->next = head; ListNode *pre_m = dummy; for(int i = 0; i < m-1; i++) pre_m = pre_m->next; ListNode *pre = pre_m->next, *cur = pre->next; for(int i = m; i < n; i++){ pre->next = cur->next; cur->next = pre_m->next; pre_m->next = cur; cur = pre->next; } return dummy->next; } };
Leetcode: Reverse Linked List II
原文：http://www.cnblogs.com/Kai-Xing/p/4103393.html