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.
解答思路：记录[m, n]范围内的结点，然后做一次reverse。
/**
* 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) {
vector<ListNode*> range(n - m + 1);
ListNode* iter = head;
for(int i = 1; i < m; ++i)
iter = iter->next;
for(int i = m, j = 0; i <= n; ++i, ++j)
{
range[j] = iter;
iter = iter->next;
}
for(size_t i = 0; i < range.size() / 2; ++i)
swap(range[i]->val, range[range.size() - i - 1]->val);
return head;
}
};
leetcode：Reverse Linked List II
原文：http://www.cnblogs.com/carsonzhu/p/4590697.html