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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.
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 };
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