标签：style blog color os io strong for div
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.
思路：先找到输入链表的第m-1个节点（即prev指向的节点）和第m个节点（即curr指向的节点），每次取curr节点的后一个节点插到prev节点之后，执行该操作n-m次即可。
1 class Solution { 2 public: 3 ListNode *reverseBetween(ListNode *head, int m, int n) { 4 if( head == 0 || head->next == 0 ) { return head; } 5 ListNode *prev = 0, *curr = head; 6 for( int i = 1; i < m; ++i ) { 7 prev = curr; curr = curr->next; 8 } 9 for( int i = m; i < n; ++i ) { 10 ListNode *tmpNode = curr->next; 11 curr->next = tmpNode->next; 12 if( prev ) { 13 tmpNode->next = prev->next; 14 prev->next = tmpNode; 15 } else { 16 tmpNode->next = head; 17 head = tmpNode; 18 } 19 } 20 return head; 21 } 22 };
Reverse Linked List II,布布扣,bubuko.com
标签：style blog color os io strong for div
原文地址：http://www.cnblogs.com/moderate-fish/p/3922925.html
