LeetCode题解： Reverse Linked List II

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：

记录[m, n]范围内的结点，然后做一次reverse。

题解：

/** * 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;    }};