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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.
/** * 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* newHead = new ListNode(-1); newHead->next = head; ListNode* p = newHead; n = n - m + 1; while(--m){ p = p->next; } ListNode* cur = p->next; ListNode* ptail = cur; while (n--) { ListNode* pNext = p->next; ListNode* curNext = cur->next; p->next = cur; cur->next = pNext; cur = curNext; ptail->next = curNext; } return newHead->next; } };
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原文地址：http://www.cnblogs.com/wxquare/p/5928092.html
