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;
}
};