LeetCode OJ - 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.

/** * 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) {        if(!head) return NULL;        ListNode *cur = head;        ListNode *front, *tail;        stack<int> st;                int start = 1;        while(cur) {            if(start == m) {                front = cur;            }            if(start >= m && start <= n) {                st.push(cur->val);            }            if(start == n) {                break;            }            start++;            cur = cur->next;        }                while(!st.empty()) {            front->val = st.top();            st.pop();            front = front->next;         }        return head;    }};