leetcode：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.

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;  
    }  
};