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leetcode - Reverse Linked List II
题目：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.
题目要求O(1)的内存消耗和只遍历一遍链表
个人思路：
1、第m-1个结点记为start，第m个结点记为before，第m+1个结点记为current，从current开始向第n个结点前进，每次都把current结点插入到start结点的后面，直到current遍历完成
2、注意一下边界情况即可
代码：
#include <stddef.h>/*struct ListNode{ int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {};};*/class Solution{public: ListNode *reverseBetween(ListNode *head, int m, int n) { if (!head) { return NULL; } //临时的头结点，仅仅是为了方便接下来的操作 ListNode *tmpHead = new ListNode(0); tmpHead->next = head; ListNode *start = tmpHead; ListNode *before = NULL; ListNode *current = head; int pos = 0; while (++pos != m) { start = current; current = current->next; } before = current; if (++pos <= n) { current = current->next; } while (pos <= n) { before->next = current->next; current->next = start->next; start->next = current; current = before->next; ++pos; } head = tmpHead->next; delete tmpHead; return head; }};
leetcode - Reverse Linked List II
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