Reverse Linked List II

Desicription

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 ≤ mn ≤ length of list.

Solution

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/**
* 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) {
ListNode* res = new ListNode(0);
res->next = head;
ListNode* cur = res;
int cnt = 0;
while(cur) {
if(cnt + 1 == m) {
ListNode* tmp = cur->next;
vector<int> vec;
while(cnt != n){
vec.push_back(tmp->val);
tmp = tmp->next, cnt++;
}
for(int i = vec.size() - 1; i >= 0; i--)
cur->next = new ListNode(vec[i]), cur = cur->next;
cur->next = tmp;
return res->next;
}
cur = cur->next, cnt++;
}
}
};